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Question: Find the solution of \({{64}^{2x-5}}=4\times {{8}^{x-5}}\). A. \(\dfrac{9}{17}\) B. \(\dfrac{17}...

Find the solution of 642x−5=4×8x−5{{64}^{2x-5}}=4\times {{8}^{x-5}}.
A. 917\dfrac{9}{17}
B. 179\dfrac{17}{9}
C. 1710\dfrac{17}{10}
D. 209\dfrac{20}{9}

Explanation

Solution

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 642x−5=4×8x−5{{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 (xa)b=xab{{\left( {{x}^{a}} \right)}^{b}}={{x}^{ab}}.
On the left-hand side, we have 642x−5=(64)2x−5=(26)2x−5{{64}^{2x-5}}={{\left( 64 \right)}^{2x-5}}={{\left( {{2}^{6}} \right)}^{2x-5}}. Now we apply the theorem and get
(26)2x−5=26(2x−5)=212x−30{{\left( {{2}^{6}} \right)}^{2x-5}}={{2}^{6\left( 2x-5 \right)}}={{2}^{12x-30}}.
On the right-hand side we have 4×8x−54\times {{8}^{x-5}}. We apply same process and get
4×8x−5=22×(23)x−5=22×23(x−5)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 xa.xb=xa+b{{x}^{a}}.{{x}^{b}}={{x}^{a+b}}.
We apply that on 22×23(x−5){{2}^{2}}\times {{2}^{3\left( x-5 \right)}} and get 22×23(x−5)=22+3x−15=23x−13{{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 212x−30=23x−13{{2}^{12x-30}}={{2}^{3x-13}}.
We know that if xa=xb⇒a=b{{x}^{a}}={{x}^{b}}\Rightarrow a=b. Applying this theorem on 212x−30=23x−13{{2}^{12x-30}}={{2}^{3x-13}}, we get
212x−30=23x−13 ⇒12x−30=3x−13 \begin{aligned} & {{2}^{12x-30}}={{2}^{3x-13}} \\\ & \Rightarrow 12x-30=3x-13 \\\ \end{aligned}
We solve the linear equation to get the value of x.
12x−30=3x−13 ⇒9x=30−13=17 ⇒x=179 \begin{aligned} & 12x-30=3x-13 \\\ & \Rightarrow 9x=30-13=17 \\\ & \Rightarrow x=\dfrac{17}{9} \\\ \end{aligned}
Therefore, the value of x is 179\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.