Question

How do you solve ${{125}^{9x-2}}=150$ ?

Answer 1

Hint: In this problem, we have to use the laws of exponents and logarithmic properties to solve the equation. Initial steps will be taken by making the base same of both the sides. Laws of exponents which will be used are:
$\begin{align}
  & \Rightarrow \dfrac{{{a}^{x}}}{{{a}^{y}}}={{a}^{x-y}} \\
 & \Rightarrow {{\left( {{a}^{x}} \right)}^{y}}={{a}^{xy}} \\
\end{align}$

Complete step-by-step answer:
Let’s solve the problem now.
We already know that exponents can be expressed in the form: ${{a}^{x}}$ and can be read as ‘a’ raise to the power ‘x’. Here, ‘a’ is the base and ‘x’ is the power. The value of ‘a’ should be greater than zero and cannot be equal to one and the value of ‘x’ can be any real number.
Now let’s discuss some important laws for exponents.
$\begin{align}
  & \Rightarrow {{a}^{x}}\times {{a}^{y}}={{a}^{x+y}} \\
 & \Rightarrow \dfrac{{{a}^{x}}}{{{a}^{y}}}={{a}^{x-y}} \\
 & \Rightarrow {{\left( {{a}^{x}} \right)}^{y}}={{a}^{xy}} \\
 & \Rightarrow {{a}^{x}}\times {{b}^{x}}={{\left( ab \right)}^{x}} \\
 & \Rightarrow \dfrac{{{a}^{x}}}{{{b}^{x}}}={{\left( \dfrac{a}{b} \right)}^{x}} \\
 & \Rightarrow {{a}^{0}}=1 \\
 & \Rightarrow {{a}^{-x}}=\dfrac{1}{{{a}^{x}}} \\
\end{align}$
First, write the equation from the question.
$\Rightarrow {{125}^{9x-2}}=150$
Express 125 and 150 in the form of exponents.
Take LCM of 125:
$\begin{align}
  & 5\left| \!{\underline {\,
  125 \,}} \right. \\
 & 5\left| \!{\underline {\,
  25 \,}} \right. \\
 & 5\left| \!{\underline {\,
  5 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
LCM of 125: $5\times 5\times 5={{5}^{3}}$
Take LCM of 150:
$\begin{align}
  & 5\left| \!{\underline {\,
  150 \,}} \right. \\
 & 5\left| \!{\underline {\,
  30 \,}} \right. \\
 & 3\left| \!{\underline {\,
  6 \,}} \right. \\
 & 2\left| \!{\underline {\,
  2 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
LCM of 150: $5\times 5\times 3\times 2=6\left( {{5}^{2}} \right)$
Change the equation by placing these exponents:
$\Rightarrow {{\left( {{5}^{3}} \right)}^{9x-2}}=6\left( {{5}^{2}} \right)$
Now, apply ${{\left( {{a}^{x}} \right)}^{y}}={{a}^{xy}}$ in above equation:
$\begin{align}
  & \Rightarrow {{5}^{3\times \left( 9x-2 \right)}}=6\left( {{5}^{2}} \right) \\
 & \Rightarrow {{5}^{27x-6}}=6\left( {{5}^{2}} \right) \\
\end{align}$
Take ${{5}^{2}}$ in the denominator of other side:
$\Rightarrow \dfrac{{{5}^{27x-6}}}{\left( {{5}^{2}} \right)}=6$
Apply the law $\Rightarrow \dfrac{{{a}^{x}}}{{{a}^{y}}}={{a}^{x-y}}$ in above equation:
$\Rightarrow {{5}^{27x-6-2}}=6\Leftrightarrow {{5}^{27x-8}}=6$
Let’s undo the exponentiation by taking the logarithm with base 5 of both sides.
$\Rightarrow {{\log }_{5}}\left( {{5}^{27x-8}} \right)={{\log }_{5}}6$
This will undo the exponent and equation changes:
$\Rightarrow 27x-8={{\log }_{5}}6$
Now, keep ‘x’ alone and take all the terms on the other side one by one:
$\begin{align}
  & \Rightarrow 27x={{\log }_{5}}6+8 \\
 & \therefore x=\dfrac{{{\log }_{5}}6+8}{27} \\
\end{align}$
So, this is the final answer.

Note: You should remember all the laws of exponents before solving such questions. After taking LCM, we took 5 in pairs because we needed 5 in exponential form. After the final step above, there is no need to solve further. Marks will not be deducted. Take care of the signs while dealing with integers.