Question

Solve for x:
\[{{\log }_{\dfrac{3}{4}}}{{\log }_{8}}({{x}^{2}}+7)+{{\log }_{\dfrac{1}{2}}}{{\log }_{\dfrac{1}{4}}}{{({{x}^{2}}+7)}^{-1}}=-2\]
This question has a multiple correct option
A). \[x=3\]
B). \[x=5\]
C). \[x=-3\]
D). \[x=6\]

Answer 1

Hint: To solve this problem, first of all we need to use the two properties of logarithm as \[{{\log }_{{{b}^{p}}}}{{a}^{n}}=\dfrac{n}{p}{{\log }_{b}}a\] and \[\log (a)-\log (b)=\log \left( \dfrac{a}{b} \right)\] then for simplicity assign variable that is \[{{\log }_{2}}({{x}^{2}}+7)=t\] by further solving this we get the final answer in terms of t and again substituting the value of t in the final answer and the find the value of x.

Complete step-by-step solution:
According to the question we have to solve for x.
Equation is given by
\[\Rightarrow {{\log }_{\dfrac{3}{4}}}{{\log }_{8}}({{x}^{2}}+7)+{{\log }_{\dfrac{1}{2}}}{{\log }_{\dfrac{1}{4}}}{{({{x}^{2}}+7)}^{-1}}=-2\]
To make the problem easier first of all we have to write in the form of powers that is
\[\Rightarrow {{\log }_{\dfrac{3}{4}}}{{\log }_{{{2}^{3}}}}({{x}^{2}}+7)+{{\log }_{\dfrac{1}{2}}}{{\log }_{{{\left( \dfrac{1}{2} \right)}^{2}}}}{{({{x}^{2}}+7)}^{-1}}=-2\]
By further simplifying this we get:
\[\Rightarrow {{\log }_{\dfrac{3}{4}}}{{\log }_{{{2}^{3}}}}({{x}^{2}}+7)+{{\log }_{\dfrac{1}{2}}}\left( {{\log }_{{{(2)}^{-2}}}}{{({{x}^{2}}+7)}^{-1}} \right)=-2\]
We will use the property of logarithmic as \[{{\log }_{{{b}^{p}}}}{{a}^{n}}=\dfrac{n}{p}{{\log }_{b}}a\] then we will get:
\[\Rightarrow {{\log }_{\dfrac{3}{4}}}\left( \dfrac{1}{3}{{\log }_{2}}({{x}^{2}}+7) \right)+{{\log }_{\dfrac{1}{2}}}\left( \dfrac{-1}{-2}{{\log }_{2}}({{x}^{2}}+7) \right)=-2\]
Now for simplicity we will be consider the variable as “t” that means \[{{\log }_{2}}({{x}^{2}}+7)=t\]
Now, substituting this value on above equation we get:
\[\Rightarrow {{\log }_{\dfrac{3}{4}}}\left( \dfrac{t}{3} \right)+{{\log }_{\dfrac{1}{2}}}\left( \dfrac{t}{2} \right)=-2\]
By using the property of logarithmic that is \[{{\log }_{b}}(a)=\dfrac{\log (a)}{\log (b)}\]then we get:
\[\Rightarrow \dfrac{\log \left( \dfrac{t}{3} \right)}{\log \left( \dfrac{3}{4} \right)}+\dfrac{\log \left( \dfrac{t}{2} \right)}{\log \left( \dfrac{1}{2} \right)}=-2\]
By using the property of logarithmic that is \[\log (a)-\log (b)=\log \left( \dfrac{a}{b} \right)\] then we get:
\[\Rightarrow \dfrac{\log \left( t \right)-\log \left( 3 \right)}{\log \left( 3 \right)-\log \left( 4 \right)}+\dfrac{\log \left( t \right)-\log \left( 2 \right)}{\log \left( 1 \right)-\log \left( 2 \right)}=-2\]
As we know that \[\log (1)=0\]
\[\Rightarrow \dfrac{\log \left( t \right)-\log \left( 3 \right)}{\log \left( 3 \right)-\log \left( 4 \right)}-\dfrac{\log \left( t \right)-\log \left( 2 \right)}{\log \left( 2 \right)}=-2\]
By rearranging the term, we get:
\[\Rightarrow \dfrac{\log \left( t \right)-\log \left( 3 \right)}{\log \left( 3 \right)-\log \left( 4 \right)}+\dfrac{\log \left( 2 \right)-\log \left( t \right)}{\log \left( 2 \right)}=-2\]
By splitting the term, we get:
\[\Rightarrow \dfrac{\log \left( t \right)}{\log \left( 3 \right)-\log \left( 4 \right)}-\dfrac{\log \left( 3 \right)}{\log \left( 3 \right)-\log \left( 4 \right)}+\dfrac{\log \left( 2 \right)}{\log \left( 2 \right)}-\dfrac{\log \left( t \right)}{\log \left( 2 \right)}=-2\]
Here, by further solving this,
\[\Rightarrow \dfrac{\log \left( t \right)}{\log \left( 3 \right)-\log \left( 4 \right)}-\dfrac{\log \left( 3 \right)}{\log \left( 3 \right)-\log \left( 4 \right)}+1-\dfrac{\log \left( t \right)}{\log \left( 2 \right)}=-2\]
By simplifying further, we get:
\[\Rightarrow \dfrac{\log \left( t \right)}{\log \left( 3 \right)-\log \left( 4 \right)}-\dfrac{\log \left( 3 \right)}{\log \left( 3 \right)-\log \left( 4 \right)}-\dfrac{\log \left( t \right)}{\log \left( 2 \right)}=-3\]
Again, rearranging the term, we get:
\[\Rightarrow \dfrac{\log \left( t \right)}{\log \left( 3 \right)-\log \left( 4 \right)}-\dfrac{\log \left( t \right)}{\log \left( 2 \right)}=-3+\dfrac{\log \left( 3 \right)}{\log \left( 3 \right)-\log \left( 4 \right)}\]
By taking LCM we get:
\[\Rightarrow \log \left( t \right)\left[ \dfrac{\log \left( 2 \right)-\log \left( 3 \right)+\log \left( 4 \right)}{\left( \log \left( 3 \right)-\log \left( 4 \right) \right)\log \left( 2 \right)} \right]=-3+\dfrac{\log \left( 3 \right)}{\log \left( 3 \right)-\log \left( 4 \right)}\]
By using the formula that is \[\log {{a}^{m}}=m\log a\] we get:
\[\Rightarrow \log \left( t \right)\left[ \dfrac{3\log \left( 2 \right)-\log \left( 3 \right)}{\left( \log \left( 3 \right)-2\log \left( 2 \right) \right)\log \left( 2 \right)} \right]=-3+\dfrac{\log \left( 3 \right)}{\log \left( 3 \right)-2\log \left( 2 \right)}\]
By cross multiplying this above equation we get:
\[\Rightarrow \log \left( t \right)\left[ \dfrac{3\log \left( 2 \right)-\log \left( 3 \right)}{\left( \log \left( 3 \right)-2\log \left( 2 \right) \right)\log \left( 2 \right)} \right]=\dfrac{-3\left( \log \left( 3 \right)-2\log \left( 2 \right) \right)+\log \left( 3 \right)}{\log \left( 3 \right)-2\log \left( 2 \right)}\]
By cancelling the term \[\log \left( 3 \right)-2\log \left( 2 \right)\] we get:
\[\Rightarrow \log \left( t \right)\left[ \dfrac{3\log \left( 2 \right)-\log \left( 3 \right)}{\log \left( 2 \right)} \right]=-3\left( \log \left( 3 \right)-2\log \left( 2 \right) \right)+\log \left( 3 \right)\]
By simplifying on RHS we get:
\[\Rightarrow \log \left( t \right)\left[ \dfrac{3\log \left( 2 \right)-\log \left( 3 \right)}{\log \left( 2 \right)} \right]=6\log \left( 2 \right)-2\log \left( 3 \right)\]
Take the 2 common on RHS we get:
\[\Rightarrow \log \left( t \right)\left[ \dfrac{3\log \left( 2 \right)-\log \left( 3 \right)}{\log \left( 2 \right)} \right]=2\left[ 3\log \left( 2 \right)-\log \left( 3 \right) \right]\]
Now, \[3\log \left( 2 \right)-\log \left( 3 \right)\] get cancelled on both sides we get:
\[\Rightarrow \log \left( t \right)\left[ \dfrac{1}{\log \left( 2 \right)} \right]=2\]
Now, multiply \[\log \left( 2 \right)\] on both sides we get:
\[\Rightarrow \log \left( t \right)=2\log (2)\]
By using the formula that is \[\log {{a}^{m}}=m\log a\]we get:
\[\Rightarrow \log \left( t \right)=\log ({{2}^{2}})=\log (4)\]
By comparing this log, we get:
\[\Rightarrow t=4\]
Now, we have to substitute the value of t,
\[\therefore {{\log }_{2}}({{x}^{2}}+7)=4\]
We know that the antilogarithmic of \[{{\log }_{a}}x={{a}^{x}}\]. So, we can take the antilogarithmic on both sides we get:
\[\Rightarrow {{2}^{{{\log }_{2}}({{x}^{2}}+7)}}={{2}^{4}}\]
We know that \[{{a}^{{{\log }_{a}}x}}=x\]
\[\Rightarrow ({{x}^{2}}+7)=16\]
by adding \[-7\] on both sides we get:
\[\Rightarrow {{x}^{2}}+7-7=16-7\]
By simplifying this we get:
\[\Rightarrow {{x}^{2}}=9\]
By squaring on both sides, we get:
\[\Rightarrow x=\pm 3\]
Therefore, value of x has two values that is \[x=3\] and \[x=-3\]
So, the correct option is “option A and option C”.

Note: In this type of problem, always remember the property and formula which we use in the problems of logarithmic. The key point of solving this problem is base changing property as we have different bases in a given log so without using this property we can’t get our answer. The logarithm rules can be used for fast exponent calculation using multiplication operations. Students should make use of the appropriate formula of logarithms wherever needed and solve the problem. In mathematics, if the base value in the logarithm function is not written, then the base is e.