Question

Find the solution of ${{64}^{2x-5}}=4\times {{8}^{x-5}}$.
A. $\dfrac{9}{17}$
B. $\dfrac{17}{9}$
C. $\dfrac{17}{10}$
D. $\dfrac{20}{9}$

Answer 1

Hint: We first need to find the proper bases that we need to use to convert the main equation. We use different theorems of indices to get the proper power value. We equate the power values to find the solution for x. We need to convert the given bases in such a way that the simplified form has an equal and basic base.

Complete step-by-step solution
We have been given an equation of indices ${{64}^{2x-5}}=4\times {{8}^{x-5}}$. The base values are 64, 4, 8.
we try to convert all the base values in the expression of 2.
We have the indices formula of ${{\left( {{x}^{a}} \right)}^{b}}={{x}^{ab}}$.
On the left-hand side, we have ${{64}^{2x-5}}={{\left( 64 \right)}^{2x-5}}={{\left( {{2}^{6}} \right)}^{2x-5}}$. Now we apply the theorem and get
${{\left( {{2}^{6}} \right)}^{2x-5}}={{2}^{6\left( 2x-5 \right)}}={{2}^{12x-30}}$.
On the right-hand side we have $4\times {{8}^{x-5}}$. We apply same process and get
$4\times {{8}^{x-5}}={{2}^{2}}\times {{\left( {{2}^{3}} \right)}^{x-5}}={{2}^{2}}\times {{2}^{3\left( x-5 \right)}}$.
We also have the theorem that ${{x}^{a}}.{{x}^{b}}={{x}^{a+b}}$.
We apply that on ${{2}^{2}}\times {{2}^{3\left( x-5 \right)}}$ and get ${{2}^{2}}\times {{2}^{3\left( x-5 \right)}}={{2}^{2+3x-15}}={{2}^{3x-13}}$.
So, we get the new form of the equation as ${{2}^{12x-30}}={{2}^{3x-13}}$.
We know that if ${{x}^{a}}={{x}^{b}}\Rightarrow a=b$. Applying this theorem on ${{2}^{12x-30}}={{2}^{3x-13}}$, we get
$\begin{align}
  & {{2}^{12x-30}}={{2}^{3x-13}} \\
 & \Rightarrow 12x-30=3x-13 \\
\end{align}$
We solve the linear equation to get the value of x.
$\begin{align}
  & 12x-30=3x-13 \\
 & \Rightarrow 9x=30-13=17 \\
 & \Rightarrow x=\dfrac{17}{9} \\
\end{align}$
Therefore, the value of x is $\dfrac{17}{9}$. The correct option is B.

Note: Although the norm is to use the basic base rule, to find the solution we could have converted the given equation into some other base other than 2. The condition is to have the same base. So, instead of 2, we could have taken the base as of 4 or 8. The power would have been fractional in some cases but the important thing is to keep the same base form.