Question

Solve the given equation: $\log 2\left( {{4}^{x+1}}+4 \right)\log 2\left( {{4}^{x}}+1 \right)={{\log }_{\dfrac{1}{\sqrt{2}}}}\sqrt{\dfrac{1}{8}}$

Answer 1

Simplify the given equation using laws of exponents such as ${{\left( {{a}^{b}} \right)}^{c}}={{a}^{bc}}$ and ${{a}^{b}}\times {{a}^{c}}={{a}^{b+c}}$. Also, use properties of log function such as $\log \left( xy \right)=\log x+\log y$ and ${{\log }_{a}}{{a}^{b}}=b$ to find the value of x which satisfies the given equation.

Complete step-by-step answer :
We have the equation $\log 2\left( {{4}^{x+1}}+4 \right)\log 2\left( {{4}^{x}}+1 \right)={{\log }_{\dfrac{1}{\sqrt{2}}}}\sqrt{\dfrac{1}{8}}$. We have to find the value of x which satisfies the given equation.
We will simplify the equation using properties of log function and laws of exponents.
We will firstly simplify the right side of the equation.
We can write 8 as $8={{2}^{3}}$.
Thus, we have ${{\log }_{\dfrac{1}{\sqrt{2}}}}\sqrt{\dfrac{1}{8}}={{\log }_{\dfrac{1}{\sqrt{2}}}}\sqrt{\dfrac{1}{{{2}^{3}}}}={{\log }_{\dfrac{1}{\sqrt{2}}}}{{\left( \dfrac{1}{\sqrt{2}} \right)}^{3}}.....\left( 1 \right)$.
We know that ${{\log }_{a}}{{a}^{b}}=b$.
Substituting $a=\dfrac{1}{\sqrt{2}},b=3$ in the above equation, we have ${{\log }_{\dfrac{1}{\sqrt{2}}}}{{\left( \dfrac{1}{\sqrt{2}} \right)}^{3}}=3.....\left( 2 \right)$.
Substituting, equation (20 in equation (1), we have ${{\log }_{\dfrac{1}{\sqrt{2}}}}\sqrt{\dfrac{1}{8}}={{\log }_{\dfrac{1}{\sqrt{2}}}}{{\left( \dfrac{1}{\sqrt{2}} \right)}^{3}}=3.....\left( 3 \right)$
We will now simplify the left side of the equation $\log 2\left( {{4}^{x+1}}+4 \right)\log 2\left( {{4}^{x}}+1 \right)={{\log }_{\dfrac{1}{\sqrt{2}}}}\sqrt{\dfrac{1}{8}}$.
We can rewrite ${{4}^{x+1}}+4$ as ${{4}^{x+1}}+4={{4}^{x}}\times 4+4=4\left( {{4}^{x}}+1 \right)$ using ${{a}^{b}}\times {{a}^{c}}={{a}^{b+c}}$.
Thus, we have $\log 2\left( {{4}^{x+1}}+4 \right)=\log 2\times 4\left( {{4}^{x}}+1 \right)$.
We know that $4={{2}^{2}}$ and $\log {{a}^{b}}=b\log a$.
Thus, we have $\log 2\left( {{4}^{x+1}}+4 \right)=\log 2\times {{2}^{2}}\left( {{4}^{x}}+1 \right)$.
We know that ${{a}^{b}}\times {{a}^{c}}={{a}^{b+c}}$.
Thus, we have $\log 2\left( {{4}^{x+1}}+4 \right)=\log {{2}^{3}}\left( {{4}^{x}}+1 \right)$.
We know that $\log \left( {{a}^{b}} \right)=b\log a$.
Thus, we have $\log 2\left( {{4}^{x+1}}+4 \right)=\log 2\times 4\left( {{4}^{x}}+1 \right)=3\log 2\left( {{4}^{x}}+1 \right)$.
So, we have $\log 2\left( {{4}^{x+1}}+4 \right)\log 2\left( {{4}^{x}}+1 \right)=3\log 2\left( {{4}^{x}}+1 \right)\log 2\left( {{4}^{x}}+1 \right)=3{{\left( \log 2\left( {{4}^{x}}+1 \right) \right)}^{2}}.....\left( 4 \right)$.
We know that $\log 2\left( {{4}^{x+1}}+4 \right)\log 2\left( {{4}^{x}}+1 \right)={{\log }_{\dfrac{1}{\sqrt{2}}}}\sqrt{\dfrac{1}{8}}$.
Substituting equation (3) and (4) in the above equation, we have $3{{\left( \log 2\left( {{4}^{x}}+1 \right) \right)}^{2}}=3$.
Rearranging the terms, we have ${{\left( \log 2\left( {{4}^{x}}+1 \right) \right)}^{2}}=1$.
Taking the square root on both sides, we have $\log 2\left( {{4}^{x}}+1 \right)=\pm 1$.
We know that $\log x=y$ can be written as $x={{e}^{y}}$.
Thus, we can rewrite $\log 2\left( {{4}^{x}}+1 \right)=\pm 1$ as $2\left( {{4}^{x}}+1 \right)={{e}^{\pm 1}}$.
Rearranging the terms, we have ${{4}^{x}}+1=\dfrac{{{e}^{\pm 1}}}{2}$.
Thus, we have ${{4}^{x}}=\dfrac{{{e}^{\pm 1}}}{2}-1$.
Taking log on both sides, we have $\log \left( {{4}^{x}} \right)=\log \left( \dfrac{{{e}^{\pm 1}}}{2}-1 \right)=x\log 4$.
Rearranging the terms, we have $x=\dfrac{\log \left( \dfrac{{{e}^{\pm 1}}}{2}-1 \right)}{\log 4}$.
Hence, the value of x which satisfies the given equation is $x=\dfrac{\log \left( \dfrac{{{e}^{\pm 1}}}{2}-1 \right)}{\log 4}$.

Note :One must know about the laws of exponents and logarithmic properties. We can’t solve this question without using them. Also, the value of ‘x’ which satisfies the given equation is a real number.