Question

Solve the following equation: \[{9^{|3x - 1|}} = {3^{8x - 2}}\].

Answer 1

Hint: We will use the properties of exponential powers. We know, \[{a^c} = {a^d} \Rightarrow c = d\]. On the left hand side, the base is \[9\] and on the right hand side the base is \[3\]. We can write \[9\] as \[{3^2}\]. Then, we know \[{\left( {{a^c}} \right)^d} = {a^{cd}}\]. Writing \[9\] as \[{3^2}\], we will have the same base on both sides and then we will equate the powers. After equating powers, we will obtain an equation with modulus terms. We know, if \[|x| = y\] then \[x = \pm y\]. We will then make two cases and obtain two values of \[x\]from each of the equations.

Complete step by step answer:
We need to solve for \[x\] in the equation : \[{9^{|3x - 1|}} = {3^{8x - 2}}\]
Now, writing \[9\]as \[3 \times 3 = {3^2}\]in the above equation, we get
\[{\left( {{3^2}} \right)^{|3x - 1|}} = {\left( 3 \right)^{8x - 2}}\]
Now, using \[{\left( {{a^c}} \right)^d} = {a^{cd}}\], we have
\[{\left( 3 \right)^{2 \times |3x - 1|}} = {\left( 3 \right)^{8x - 2}}\]
Using exponential property \[{a^c} = {a^d} \Rightarrow c = d\], we have
\[2 \times |3x - 1| = 8x - 2\]
Now, dividing the left hand side and right hand side by \[2\]
\[\dfrac{{2 \times |3x - 1|}}{2} = \dfrac{{8x - 2}}{2}\]
Taking out \[2\]common from the Numerator of the right hand side.
\[\dfrac{{2 \times |3x - 1|}}{2} = \dfrac{{2 \times (4x - 1)}}{2}\]
Now, cancelling \[2\]from left and right hand side both, we get
\[|3x - 1| = 4x - 1\]
Now using \[|x| = y \Rightarrow x = \pm y\] in the above equation
\[3x - 1 = \pm (4x - 1)\]
We will now separate them into two cases.
CASE 1 - \[3x - 1 = + (4x - 1)\]
CASE 2 - \[3x - 1 = - (4x - 1)\]
Now, solving for \[x\] from both the equations.
From CASE 1:
\[3x - 1 = + (4x - 1)\]
\[3x - 1 = 4x - 1\]
Now, reshuffling the equations, we get
\[3x - 4x = - 1 + 1\]
Solving both Left Hand Side and Right
\[ - 1x = 0\]
\[ - x = 0\]
Now, multiplying both the side by \[ - 1\]
\[ - 1( - x) = ( - 1)0\]
\[x = 0 - - - - - (1)\]
From CASE 2:
\[3x - 1 = - (4x - 1)\]
Opening the brackets, we get
\[3x - 1 = - 4x - ( - 1)\]
\[3x - 1 = - 4x + 1\]
Now, reshuffling the terms in the above equations
\[3x + 4x = + 1 + 1\]
Solving the Right Hand Side and Left Hand Side
\[7x = 2\]
Dividing both the sides by \[7\], we get
\[\dfrac{{7x}}{7} = \dfrac{2}{7}\]
\[x = \dfrac{2}{7} - - - - - (2)\]
From (1) and (2), we get
\[x = 0\] and \[x = \dfrac{2}{7}\], which is the required answer.

Note:
We need to be very thorough with the properties of exponential powers. Unless and until we don’t make the base same, we won’t be able to solve it. And, when we are solving a modulus problem, we usually don’t consider the two cases but solve for the plus case only. Considering two cases, we need to solve them separately and then we will get two different values of \[x\]. While solving the equations, we need to be very careful with the signs.