Solving this equation: $3^{log_{4}x+frac{1}{2}}+3^{log_{4}x-frac{1}{2}}=sqrt{x}$
$begingroup$
Solve this equation:
$$3^{log_{4}x+frac{1}{2}}+3^{log_{4}x-frac{1}{2}}=sqrt{x}qquad (1)$$
I tried to make both sides of the equation have a same base and I started:
$$(1)Leftrightarrow 3^{log_{4}x}.sqrt{3}+ frac{3^{log_{4}x}}{sqrt{3}} = sqrt{x}$$
$$Leftrightarrow 3^{log_{4}x}.3+ 3^{log_{4}x} = sqrt{3x}$$
$$Leftrightarrow 4.3^{log_{4}x}= sqrt{3x}$$
At this step, I can't continue. Please help me!
logarithms
edited Dec 11 '18 at 6:45
Robert Z
101k1070143
asked Dec 11 '18 at 6:29
Trần TuấnTrần Tuấn
254
$endgroup$
add a comment |
$begingroup$
Solve this equation:
$$3^{log_{4}x+frac{1}{2}}+3^{log_{4}x-frac{1}{2}}=sqrt{x}qquad (1)$$
I tried to make both sides of the equation have a same base and I started:
$$(1)Leftrightarrow 3^{log_{4}x}.sqrt{3}+ frac{3^{log_{4}x}}{sqrt{3}} = sqrt{x}$$
$$Leftrightarrow 3^{log_{4}x}.3+ 3^{log_{4}x} = sqrt{3x}$$
$$Leftrightarrow 4.3^{log_{4}x}= sqrt{3x}$$
At this step, I can't continue. Please help me!
logarithms
edited Dec 11 '18 at 6:45
Robert Z
101k1070143
asked Dec 11 '18 at 6:29
Trần TuấnTrần Tuấn
254
$endgroup$
add a comment |
$begingroup$
Solve this equation:
$$3^{log_{4}x+frac{1}{2}}+3^{log_{4}x-frac{1}{2}}=sqrt{x}qquad (1)$$
I tried to make both sides of the equation have a same base and I started:
$$(1)Leftrightarrow 3^{log_{4}x}.sqrt{3}+ frac{3^{log_{4}x}}{sqrt{3}} = sqrt{x}$$
$$Leftrightarrow 3^{log_{4}x}.3+ 3^{log_{4}x} = sqrt{3x}$$
$$Leftrightarrow 4.3^{log_{4}x}= sqrt{3x}$$
At this step, I can't continue. Please help me!
logarithms
edited Dec 11 '18 at 6:45
Robert Z
101k1070143
asked Dec 11 '18 at 6:29
Trần TuấnTrần Tuấn
254
$endgroup$
Solve this equation:
$$3^{log_{4}x+frac{1}{2}}+3^{log_{4}x-frac{1}{2}}=sqrt{x}qquad (1)$$
I tried to make both sides of the equation have a same base and I started:
$$(1)Leftrightarrow 3^{log_{4}x}.sqrt{3}+ frac{3^{log_{4}x}}{sqrt{3}} = sqrt{x}$$
$$Leftrightarrow 3^{log_{4}x}.3+ 3^{log_{4}x} = sqrt{3x}$$
$$Leftrightarrow 4.3^{log_{4}x}= sqrt{3x}$$
At this step, I can't continue. Please help me!
logarithms
logarithms
edited Dec 11 '18 at 6:45
Robert Z
101k1070143
asked Dec 11 '18 at 6:29
Trần TuấnTrần Tuấn
254
edited Dec 11 '18 at 6:45
Robert Z
101k1070143
asked Dec 11 '18 at 6:29
Trần TuấnTrần Tuấn
254
edited Dec 11 '18 at 6:45
Robert Z
101k1070143
edited Dec 11 '18 at 6:45
Robert Z
101k1070143
edited Dec 11 '18 at 6:45
Robert Z
101k1070143
101k1070143
asked Dec 11 '18 at 6:29
Trần TuấnTrần Tuấn
254
asked Dec 11 '18 at 6:29
Trần TuấnTrần Tuấn
254
asked Dec 11 '18 at 6:29
Trần TuấnTrần Tuấn
254
254
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
First of all impose the necessary existence conditions, that is: $x > 0$ for the logarithms and $x geq 0$ for the square root.
That is, eventually,
$$x > 0$$
for the whole equation.
Then follow Siong Thye Goh reasoning, obtaining the final equation he wrote.
At that point:
$$1 + log_4(3)log_4(x) = frac{log_4(3)}{2} + frac{log_4(x)}{2}$$
$$log_4(x)left(log_4(3) - frac{1}{2}right) = frac{log_4(3)}{2} - 1$$
$$log_4(x) = frac{frac{log_4(3)}{2} - 1}{log_4(3) - frac{1}{2}} = frac{log_4(3)-2}{2log_4(3)-1}$$
To solve for $x$ take the exponential base 4 of both terms, getting:
$$large x = large 4^{frac{log_4(3)-2}{2log_4(3)-1}}$$
answered Dec 11 '18 at 6:49
Von NeumannVon Neumann
16.5k72545
$endgroup$
add a comment |
$begingroup$
Guide:
Taking $log_4$ on both sides,
$$1 + log_4 3 cdot log_4 x = frac12 (log_43 + log_4 x)$$
Solve for $log_4 x$.
answered Dec 11 '18 at 6:32
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
$begingroup$
You may also continue as follows:
$$begin{eqnarray*}
4 cdot 3^{2 cdot log_{4}sqrt{x}} &= & sqrt{3}sqrt{x} Leftrightarrow \
4 cdot 9^{log_{4}sqrt{x}} &= & sqrt{3}sqrt{x} Leftrightarrow \
4 cdot 4^{log_4{9} cdot log_{4}sqrt{x}} &= & sqrt{3}sqrt{x} Leftrightarrow \
(sqrt{x})^{log_4{9} -1} &= & frac{sqrt{3}}{4} Leftrightarrow\
x & = & left( frac{sqrt{3}}{4}right)^{frac{log_4{9}-1}{2}} approx 0.0571725
end{eqnarray*}$$
answered Dec 11 '18 at 6:56
trancelocationtrancelocation
13.2k1827
$endgroup$
add a comment |
$begingroup$
Let's generalise a bit, with parameters say $a, b in (0, infty)setminus {1}$ subject to $a^2 neq b$ and let's try to solve the equation:
$$a^{log_{b}x+frac{1}{2}}+a^{log_{b}x-frac{1}{2}}=sqrt{x}$$
Notice that the left-hand side can be rewritten as
$$a^{log_{b}x}(sqrt{a}+frac{1}{sqrt{a}})=b^{log_{b}acdot log_{b}x}(sqrt{a}+frac{1}{sqrt{a}})=x^{log_{b}a}(sqrt{a}+frac{1}{sqrt{a}})$$
As you are dealing exclusively with strictly positive reals, your given equation is equivalent to its square, so to speak:
$$frac{(a+1)^2}{a}x^{2log_{b}a}=x$$
which leads to
$$x^{2log_{b}a-1}=frac{a}{(a+1)^2}$$
Since the right-hand side is never $1$ (you can try to see why), this is why we initially imposed the relation of inequality between $a$ and $b$; it is satisfied in the particular case of your equation.
We finally have the solution:
$$x=left(frac{a}{(a+1)^2}right)^{frac{1}{2log_{b}a-1}}$$
answered Dec 11 '18 at 7:02
ΑΘΩΑΘΩ
3436
$endgroup$
add a comment |
$begingroup$
Same solution with same approach
But simplified with substitution
answered Dec 11 '18 at 7:24
user579689user579689
113
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First of all impose the necessary existence conditions, that is: $x > 0$ for the logarithms and $x geq 0$ for the square root.
That is, eventually,
$$x > 0$$
for the whole equation.
Then follow Siong Thye Goh reasoning, obtaining the final equation he wrote.
At that point:
$$1 + log_4(3)log_4(x) = frac{log_4(3)}{2} + frac{log_4(x)}{2}$$
$$log_4(x)left(log_4(3) - frac{1}{2}right) = frac{log_4(3)}{2} - 1$$
$$log_4(x) = frac{frac{log_4(3)}{2} - 1}{log_4(3) - frac{1}{2}} = frac{log_4(3)-2}{2log_4(3)-1}$$
To solve for $x$ take the exponential base 4 of both terms, getting:
$$large x = large 4^{frac{log_4(3)-2}{2log_4(3)-1}}$$
answered Dec 11 '18 at 6:49
Von NeumannVon Neumann
16.5k72545
$endgroup$
add a comment |
$begingroup$
First of all impose the necessary existence conditions, that is: $x > 0$ for the logarithms and $x geq 0$ for the square root.
That is, eventually,
$$x > 0$$
for the whole equation.
Then follow Siong Thye Goh reasoning, obtaining the final equation he wrote.
At that point:
$$1 + log_4(3)log_4(x) = frac{log_4(3)}{2} + frac{log_4(x)}{2}$$
$$log_4(x)left(log_4(3) - frac{1}{2}right) = frac{log_4(3)}{2} - 1$$
$$log_4(x) = frac{frac{log_4(3)}{2} - 1}{log_4(3) - frac{1}{2}} = frac{log_4(3)-2}{2log_4(3)-1}$$
To solve for $x$ take the exponential base 4 of both terms, getting:
$$large x = large 4^{frac{log_4(3)-2}{2log_4(3)-1}}$$
answered Dec 11 '18 at 6:49
Von NeumannVon Neumann
16.5k72545
$endgroup$
add a comment |
$begingroup$
First of all impose the necessary existence conditions, that is: $x > 0$ for the logarithms and $x geq 0$ for the square root.
That is, eventually,
$$x > 0$$
for the whole equation.
Then follow Siong Thye Goh reasoning, obtaining the final equation he wrote.
At that point:
$$1 + log_4(3)log_4(x) = frac{log_4(3)}{2} + frac{log_4(x)}{2}$$
$$log_4(x)left(log_4(3) - frac{1}{2}right) = frac{log_4(3)}{2} - 1$$
$$log_4(x) = frac{frac{log_4(3)}{2} - 1}{log_4(3) - frac{1}{2}} = frac{log_4(3)-2}{2log_4(3)-1}$$
To solve for $x$ take the exponential base 4 of both terms, getting:
$$large x = large 4^{frac{log_4(3)-2}{2log_4(3)-1}}$$
answered Dec 11 '18 at 6:49
Von NeumannVon Neumann
16.5k72545
$endgroup$
First of all impose the necessary existence conditions, that is: $x > 0$ for the logarithms and $x geq 0$ for the square root.
That is, eventually,
$$x > 0$$
for the whole equation.
Then follow Siong Thye Goh reasoning, obtaining the final equation he wrote.
At that point:
$$1 + log_4(3)log_4(x) = frac{log_4(3)}{2} + frac{log_4(x)}{2}$$
$$log_4(x)left(log_4(3) - frac{1}{2}right) = frac{log_4(3)}{2} - 1$$
$$log_4(x) = frac{frac{log_4(3)}{2} - 1}{log_4(3) - frac{1}{2}} = frac{log_4(3)-2}{2log_4(3)-1}$$
To solve for $x$ take the exponential base 4 of both terms, getting:
$$large x = large 4^{frac{log_4(3)-2}{2log_4(3)-1}}$$
answered Dec 11 '18 at 6:49
Von NeumannVon Neumann
16.5k72545
answered Dec 11 '18 at 6:49
Von NeumannVon Neumann
16.5k72545
answered Dec 11 '18 at 6:49
Von NeumannVon Neumann
16.5k72545
answered Dec 11 '18 at 6:49
Von NeumannVon Neumann
16.5k72545
16.5k72545
add a comment |
add a comment |
$begingroup$
Guide:
Taking $log_4$ on both sides,
$$1 + log_4 3 cdot log_4 x = frac12 (log_43 + log_4 x)$$
Solve for $log_4 x$.
answered Dec 11 '18 at 6:32
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
$begingroup$
Guide:
Taking $log_4$ on both sides,
$$1 + log_4 3 cdot log_4 x = frac12 (log_43 + log_4 x)$$
Solve for $log_4 x$.
answered Dec 11 '18 at 6:32
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
$begingroup$
Guide:
Taking $log_4$ on both sides,
$$1 + log_4 3 cdot log_4 x = frac12 (log_43 + log_4 x)$$
Solve for $log_4 x$.
answered Dec 11 '18 at 6:32
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
Guide:
Taking $log_4$ on both sides,
$$1 + log_4 3 cdot log_4 x = frac12 (log_43 + log_4 x)$$
Solve for $log_4 x$.
answered Dec 11 '18 at 6:32
Siong Thye GohSiong Thye Goh
103k1468119
answered Dec 11 '18 at 6:32
Siong Thye GohSiong Thye Goh
103k1468119
answered Dec 11 '18 at 6:32
Siong Thye GohSiong Thye Goh
103k1468119
answered Dec 11 '18 at 6:32
Siong Thye GohSiong Thye Goh
103k1468119
103k1468119
add a comment |
add a comment |
$begingroup$
You may also continue as follows:
$$begin{eqnarray*}
4 cdot 3^{2 cdot log_{4}sqrt{x}} &= & sqrt{3}sqrt{x} Leftrightarrow \
4 cdot 9^{log_{4}sqrt{x}} &= & sqrt{3}sqrt{x} Leftrightarrow \
4 cdot 4^{log_4{9} cdot log_{4}sqrt{x}} &= & sqrt{3}sqrt{x} Leftrightarrow \
(sqrt{x})^{log_4{9} -1} &= & frac{sqrt{3}}{4} Leftrightarrow\
x & = & left( frac{sqrt{3}}{4}right)^{frac{log_4{9}-1}{2}} approx 0.0571725
end{eqnarray*}$$
answered Dec 11 '18 at 6:56
trancelocationtrancelocation
13.2k1827
$endgroup$
add a comment |
$begingroup$
You may also continue as follows:
$$begin{eqnarray*}
4 cdot 3^{2 cdot log_{4}sqrt{x}} &= & sqrt{3}sqrt{x} Leftrightarrow \
4 cdot 9^{log_{4}sqrt{x}} &= & sqrt{3}sqrt{x} Leftrightarrow \
4 cdot 4^{log_4{9} cdot log_{4}sqrt{x}} &= & sqrt{3}sqrt{x} Leftrightarrow \
(sqrt{x})^{log_4{9} -1} &= & frac{sqrt{3}}{4} Leftrightarrow\
x & = & left( frac{sqrt{3}}{4}right)^{frac{log_4{9}-1}{2}} approx 0.0571725
end{eqnarray*}$$
answered Dec 11 '18 at 6:56
trancelocationtrancelocation
13.2k1827
$endgroup$
add a comment |
$begingroup$
You may also continue as follows:
$$begin{eqnarray*}
4 cdot 3^{2 cdot log_{4}sqrt{x}} &= & sqrt{3}sqrt{x} Leftrightarrow \
4 cdot 9^{log_{4}sqrt{x}} &= & sqrt{3}sqrt{x} Leftrightarrow \
4 cdot 4^{log_4{9} cdot log_{4}sqrt{x}} &= & sqrt{3}sqrt{x} Leftrightarrow \
(sqrt{x})^{log_4{9} -1} &= & frac{sqrt{3}}{4} Leftrightarrow\
x & = & left( frac{sqrt{3}}{4}right)^{frac{log_4{9}-1}{2}} approx 0.0571725
end{eqnarray*}$$
answered Dec 11 '18 at 6:56
trancelocationtrancelocation
13.2k1827
$endgroup$
You may also continue as follows:
$$begin{eqnarray*}
4 cdot 3^{2 cdot log_{4}sqrt{x}} &= & sqrt{3}sqrt{x} Leftrightarrow \
4 cdot 9^{log_{4}sqrt{x}} &= & sqrt{3}sqrt{x} Leftrightarrow \
4 cdot 4^{log_4{9} cdot log_{4}sqrt{x}} &= & sqrt{3}sqrt{x} Leftrightarrow \
(sqrt{x})^{log_4{9} -1} &= & frac{sqrt{3}}{4} Leftrightarrow\
x & = & left( frac{sqrt{3}}{4}right)^{frac{log_4{9}-1}{2}} approx 0.0571725
end{eqnarray*}$$
answered Dec 11 '18 at 6:56
trancelocationtrancelocation
13.2k1827
answered Dec 11 '18 at 6:56
trancelocationtrancelocation
13.2k1827
answered Dec 11 '18 at 6:56
trancelocationtrancelocation
13.2k1827
answered Dec 11 '18 at 6:56
trancelocationtrancelocation
13.2k1827
13.2k1827
add a comment |
add a comment |
$begingroup$
Let's generalise a bit, with parameters say $a, b in (0, infty)setminus {1}$ subject to $a^2 neq b$ and let's try to solve the equation:
$$a^{log_{b}x+frac{1}{2}}+a^{log_{b}x-frac{1}{2}}=sqrt{x}$$
Notice that the left-hand side can be rewritten as
$$a^{log_{b}x}(sqrt{a}+frac{1}{sqrt{a}})=b^{log_{b}acdot log_{b}x}(sqrt{a}+frac{1}{sqrt{a}})=x^{log_{b}a}(sqrt{a}+frac{1}{sqrt{a}})$$
As you are dealing exclusively with strictly positive reals, your given equation is equivalent to its square, so to speak:
$$frac{(a+1)^2}{a}x^{2log_{b}a}=x$$
which leads to
$$x^{2log_{b}a-1}=frac{a}{(a+1)^2}$$
Since the right-hand side is never $1$ (you can try to see why), this is why we initially imposed the relation of inequality between $a$ and $b$; it is satisfied in the particular case of your equation.
We finally have the solution:
$$x=left(frac{a}{(a+1)^2}right)^{frac{1}{2log_{b}a-1}}$$
answered Dec 11 '18 at 7:02
ΑΘΩΑΘΩ
3436
$endgroup$
add a comment |
$begingroup$
Let's generalise a bit, with parameters say $a, b in (0, infty)setminus {1}$ subject to $a^2 neq b$ and let's try to solve the equation:
$$a^{log_{b}x+frac{1}{2}}+a^{log_{b}x-frac{1}{2}}=sqrt{x}$$
Notice that the left-hand side can be rewritten as
$$a^{log_{b}x}(sqrt{a}+frac{1}{sqrt{a}})=b^{log_{b}acdot log_{b}x}(sqrt{a}+frac{1}{sqrt{a}})=x^{log_{b}a}(sqrt{a}+frac{1}{sqrt{a}})$$
As you are dealing exclusively with strictly positive reals, your given equation is equivalent to its square, so to speak:
$$frac{(a+1)^2}{a}x^{2log_{b}a}=x$$
which leads to
$$x^{2log_{b}a-1}=frac{a}{(a+1)^2}$$
Since the right-hand side is never $1$ (you can try to see why), this is why we initially imposed the relation of inequality between $a$ and $b$; it is satisfied in the particular case of your equation.
We finally have the solution:
$$x=left(frac{a}{(a+1)^2}right)^{frac{1}{2log_{b}a-1}}$$
answered Dec 11 '18 at 7:02
ΑΘΩΑΘΩ
3436
$endgroup$
add a comment |
$begingroup$
Let's generalise a bit, with parameters say $a, b in (0, infty)setminus {1}$ subject to $a^2 neq b$ and let's try to solve the equation:
$$a^{log_{b}x+frac{1}{2}}+a^{log_{b}x-frac{1}{2}}=sqrt{x}$$
Notice that the left-hand side can be rewritten as
$$a^{log_{b}x}(sqrt{a}+frac{1}{sqrt{a}})=b^{log_{b}acdot log_{b}x}(sqrt{a}+frac{1}{sqrt{a}})=x^{log_{b}a}(sqrt{a}+frac{1}{sqrt{a}})$$
As you are dealing exclusively with strictly positive reals, your given equation is equivalent to its square, so to speak:
$$frac{(a+1)^2}{a}x^{2log_{b}a}=x$$
which leads to
$$x^{2log_{b}a-1}=frac{a}{(a+1)^2}$$
Since the right-hand side is never $1$ (you can try to see why), this is why we initially imposed the relation of inequality between $a$ and $b$; it is satisfied in the particular case of your equation.
We finally have the solution:
$$x=left(frac{a}{(a+1)^2}right)^{frac{1}{2log_{b}a-1}}$$
answered Dec 11 '18 at 7:02
ΑΘΩΑΘΩ
3436
$endgroup$
Let's generalise a bit, with parameters say $a, b in (0, infty)setminus {1}$ subject to $a^2 neq b$ and let's try to solve the equation:
$$a^{log_{b}x+frac{1}{2}}+a^{log_{b}x-frac{1}{2}}=sqrt{x}$$
Notice that the left-hand side can be rewritten as
$$a^{log_{b}x}(sqrt{a}+frac{1}{sqrt{a}})=b^{log_{b}acdot log_{b}x}(sqrt{a}+frac{1}{sqrt{a}})=x^{log_{b}a}(sqrt{a}+frac{1}{sqrt{a}})$$
As you are dealing exclusively with strictly positive reals, your given equation is equivalent to its square, so to speak:
$$frac{(a+1)^2}{a}x^{2log_{b}a}=x$$
which leads to
$$x^{2log_{b}a-1}=frac{a}{(a+1)^2}$$
Since the right-hand side is never $1$ (you can try to see why), this is why we initially imposed the relation of inequality between $a$ and $b$; it is satisfied in the particular case of your equation.
We finally have the solution:
$$x=left(frac{a}{(a+1)^2}right)^{frac{1}{2log_{b}a-1}}$$
answered Dec 11 '18 at 7:02
ΑΘΩΑΘΩ
3436
answered Dec 11 '18 at 7:02
ΑΘΩΑΘΩ
3436
answered Dec 11 '18 at 7:02
ΑΘΩΑΘΩ
3436
answered Dec 11 '18 at 7:02
ΑΘΩΑΘΩ
3436
3436
add a comment |
add a comment |
$begingroup$
Same solution with same approach
But simplified with substitution
answered Dec 11 '18 at 7:24
user579689user579689
113
$endgroup$
add a comment |
$begingroup$
Same solution with same approach
But simplified with substitution
answered Dec 11 '18 at 7:24
user579689user579689
113
$endgroup$
add a comment |
$begingroup$
Same solution with same approach
But simplified with substitution
answered Dec 11 '18 at 7:24
user579689user579689
113
$endgroup$
Same solution with same approach
But simplified with substitution
answered Dec 11 '18 at 7:24
user579689user579689
113
answered Dec 11 '18 at 7:24
user579689user579689
113
answered Dec 11 '18 at 7:24
user579689user579689
113
answered Dec 11 '18 at 7:24
user579689user579689
113
113
add a comment |
add a comment |
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