Question

How do you solve ${{\left( \dfrac{1}{64} \right)}^{x}}={{2}^{3x+5}}$ ?

Answer 1

Hint: In this question, we have to find the value of x. Since the equation consists of the exponent term, thus we will apply the exponent rule to get the solution. First, we will find the LCM of 64 and then rewrite it in terms of number 2. After that, we will apply the formula $\dfrac{1}{a}={{a}^{-1}}$ on the left-hand side of the equation. Then, we will apply the exponent formula which states that if the bases are the same on both sides of the equation, then their powers are also equal to each other. After the necessary calculations, we get the value of x, which is the required solution for the problem.

Complete step by step solution:
According to the question, we have to find the value of x.
Thus we will apply the exponent rule to get the required solution.
The equation given to us is ${{\left( \dfrac{1}{64} \right)}^{x}}={{2}^{3x+5}}$ ---------- (1)
Now, we will first find the least common multiple of 64, we get
$\begin{align}
  & \text{ 2}\left| \!{\underline {\,
  64 \,}} \right. \\
 & \text{ 2}\left| \!{\underline {\,
  32 \,}} \right. \\
 & \text{ 2}\left| \!{\underline {\,
  16 \,}} \right. \\
 & \text{ 2}\left| \!{\underline {\,
  8 \,}} \right. \\
 & \text{ 2}\left| \!{\underline {\,
  4 \,}} \right. \\
 & \text{ 2}\left| \!{\underline {\,
  2 \,}} \right. \\
 & \text{ }\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
Therefore,
$(64)={{2}^{6}}$ --------- (2)
Thus, we will now substitute the value of equation (2) on the left-hand side of equation (1), we get
${{\left( \dfrac{1}{{{2}^{6}}} \right)}^{x}}={{2}^{3x+5}}$
Now, we will apply the algebraic formula ${{\left( \dfrac{1}{{{a}^{b}}} \right)}^{c}}={{\left( \dfrac{1}{a} \right)}^{bc}}$ in the above equation, we get
${{\left( \dfrac{1}{2} \right)}^{6x}}={{2}^{3x+5}}$
Now, we will apply the formula $\dfrac{1}{a}={{a}^{-1}}$ on the left-hand side in the above equation, we get
${{\left( 2 \right)}^{-6x}}={{2}^{3x+5}}$
As we know the exponent rule which states that if the bases are the same on both the sides of the equation, then their powers are also equal to each other, thus we will apply the exponent rule in the above equation as 2 is the same on both sides of the equation, we get
$-6x=3x+5$
Now, we will subtract 3x on both sides in the above equation, we get
$-6x-3x=3x+5-3x$
As we know, the same terms with opposite signs cancel out each other, thus we get
$-9x=5$
Now, we will divide 9 on both sides in the above equation, we get
$-\dfrac{9}{9}x=\dfrac{5}{9}$
On further simplification, we get
$-x=\dfrac{5}{9}$
In the last, we will multiply (-1) on both sides in the above equation, we get
$-x.(-1)=\dfrac{5}{9}.(-1)$
On further solving the above equation, we get
$x=-\dfrac{5}{9}$

Therefore, for the equation ${{\left( \dfrac{1}{64} \right)}^{x}}={{2}^{3x+5}}$ , the value of x is equal to $-\dfrac{5}{9}$

Note: While solving this problem, do mention all the formulas to avoid confusion and mathematical errors. One of the alternative methods to solve this problem is after this ${{\left( 2 \right)}^{-6x}}={{2}^{3x+5}}$ step, you can divide ${{2}^{3x+5}}$ on both sides of the equation and apply the exponent rule, that is both the bases are the same in the numerator and the denominator, then the powers are subtracted to each other. Then make the necessary calculations to get the required result for the solution.