Question

How do you solve the logarithmic equation $2{{\log }_{6}}4-\dfrac{1}{4}{{\log }_{6}}16={{\log }_{6}}x$ ?

Answer 1

To solve $2{{\log }_{6}}4-\dfrac{1}{4}{{\log }_{6}}16={{\log }_{6}}x$ , we have to write 16 in terms of 4 as ${{4}^{2}}=16$ . Then, we have to apply the logarithmic property $\log {{x}^{n}}=n\log x$ and solve. We have to take the exponents on both the side of the equation and simplify using the logarithm property ${{b}^{{{\log }_{b}}x}}=x$ and rules of exponents.

Complete step-by-step solution:
We have to solve $2{{\log }_{6}}4-\dfrac{1}{4}{{\log }_{6}}16={{\log }_{6}}x$ . Let us write 16 in terms of 4 as ${{4}^{2}}=16$ .
$\Rightarrow 2{{\log }_{6}}4-\dfrac{1}{4}{{\log }_{6}}\left( {{4}^{2}} \right)={{\log }_{6}}x$
We know that $\log {{x}^{n}}=n\log x$ . We can write the above equation as
$\Rightarrow 2{{\log }_{6}}4-\dfrac{2}{4}{{\log }_{6}}4={{\log }_{6}}x$
Let us simplify the second term of the LHS.
$\Rightarrow 2{{\log }_{6}}4-\dfrac{1}{2}{{\log }_{6}}4={{\log }_{6}}x$ .
Now, we have to simplify the like terms.
$\Rightarrow \dfrac{4{{\log }_{6}}4-{{\log }_{6}}4}{2}={{\log }_{6}}x$
Let us subtract the terms on the LHS. We can write the result of this step as
$\Rightarrow \dfrac{3{{\log }_{6}}4}{2}={{\log }_{6}}x$
We can rewrite the above equation as
$\Rightarrow \dfrac{3}{2}{{\log }_{6}}4={{\log }_{6}}x$
We know that $\log {{x}^{n}}=n\log x$ . Hence, we can write the above equation as
$\Rightarrow {{\log }_{6}}\left( {{4}^{\dfrac{3}{2}}} \right)={{\log }_{6}}x$
We have to take the exponents on both the sides by 6 since the base is 6.
$\Rightarrow {{6}^{{{\log }_{6}}\left( {{4}^{\dfrac{3}{2}}} \right)}}={{6}^{{{\log }_{6}}x}}$
We know that ${{b}^{{{\log }_{b}}x}}=x$ . Therefore, the above equation can be written as
$\Rightarrow {{4}^{\dfrac{3}{2}}}=x$
We know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ . Then, the above equation becomes

& \Rightarrow x={{\left( {{4}^{\dfrac{1}{2}}} \right)}^{3}} \\\ & \Rightarrow x={{\left( \sqrt{4} \right)}^{3}} \\\ \end{aligned}$$ We know that $\sqrt{4}=2$ . Therefore, the above result becomes $$\begin{aligned} & \Rightarrow x={{\left( 2 \right)}^{3}} \\\ & \Rightarrow x=8 \\\ \end{aligned}$$ **Therefore, the solution of $2{{\log }_{6}}4-\dfrac{1}{4}{{\log }_{6}}16={{\log }_{6}}x$ is $x=8$.** **Note:** Students must be thorough with the logarithmic rules and properties. They must know to take the exponents on both the sides. This is the main section of the solution and there is chance of making mistakes here. Students must also be thorough with the rules of exponents.