Question

Solve the following equation:
\[\begin{align}
  & \left( i \right){{5}^{x-3}}\times {{3}^{2x-8}}=225 \\
 & \left( ii \right){{\left( 25 \right)}^{x-1}}={{5}^{2x-1}}-100 \\
\end{align}\]

Answer 1

Hint: In this question, we need to find the value of x such that the left side of the equation is equal to the right side of the equation. For this, we will use the laws of exponents which are:
\[\begin{align}
  & \left( i \right){{a}^{m}}\cdot {{a}^{n}}={{a}^{m+n}} \\
 & \left( ii \right)\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}} \\
 & \left( iii \right){{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}} \\
\end{align}\]
We will try to make the right side in the same base as the left side and then compare the exponents to get the value of x.

Complete step-by-step answer:
Let us solve (1) first.
Equation is given as $\left( i \right){{5}^{x-3}}\times {{3}^{2x-8}}=225$.
We need to find the value of x. For this, let us first factorise 225. As we know, 225 can be written as $5\times 5\times 3\times 3$. So, we can also write 225 as ${{5}^{2}}\times {{3}^{2}}$ we get: ${{5}^{x-3}}\times {{3}^{2x-8}}={{5}^{2}}\cdot {{3}^{2}}$.
We know $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ so we get: $\dfrac{{{5}^{x}}}{{{5}^{3}}}\times \dfrac{{{3}^{2x}}}{{{3}^{8}}}={{5}^{2}}\cdot {{3}^{2}}$.
Cross multiplying we get: ${{5}^{x}}\times {{3}^{2x}}={{5}^{2}}\cdot {{3}^{2}}\cdot {{5}^{3}}\cdot {{3}^{8}}$.
We know that ${{a}^{m}}\cdot {{a}^{n}}={{a}^{m+n}}$ hence, we get: ${{5}^{x}}\times {{3}^{2x}}={{5}^{2+3}}\times {{3}^{2+8}}\Rightarrow {{5}^{x}}\times {{3}^{2x}}={{5}^{5}}\times {{3}^{10}}$.
Now 10 can be written as $2\times 5$ so we get ${{5}^{x}}\times {{3}^{2x}}={{5}^{5}}\times {{3}^{2\times 5}}$.
By comparing, we can see that x = 5. Hence the value of x is 5.

Solving (2).
Equation is given as $\left( ii \right){{\left( 25 \right)}^{x-1}}={{5}^{2x-1}}-100$.
Let us bring terms containing x on one side and constant on the other side, we get:
${{5}^{2x-1}}-{{\left( 25 \right)}^{x-1}}=100$.
We know that $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ so we get: $\dfrac{{{5}^{2x}}}{5}-\dfrac{{{\left( 25 \right)}^{x}}}{25}=100$.
Since ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ so we can write ${{5}^{2x}}$ as ${{\left( {{5}^{2}} \right)}^{x}}$ and value of ${{5}^{2}}$ is equal to 25. So we get: $\dfrac{{{25}^{x}}}{5}-\dfrac{{{25}^{x}}}{25}=100$.
Taking LCM of 25 on the left side of the equation, we get: $\dfrac{5\times {{25}^{x}}-{{25}^{x}}}{25}=100$.
Taking \[{{25}^{x}}\] common on the left side, we get: $\dfrac{{{25}^{x}}\left( 5-1 \right)}{25}=100\Rightarrow {{25}^{x}}\times \dfrac{4}{25}=100$.
Taking $\dfrac{4}{25}$ to other side, we get: ${{25}^{x}}=\dfrac{100\times 25}{4}\Rightarrow {{25}^{x}}=25\times 25$.
Now $25\times 25$ can be written as ${{\left( 25 \right)}^{2}}$ so we get: ${{25}^{x}}={{\left( 25 \right)}^{2}}$.
By comparing, we can see that x is equal to 2.
Hence the required value of x is 2.

Note: Students should take care of the signs while solving these sums. Try to simplify the left side of the equation as much as possible before comparing. Students should note that, base should be the same while comparing. Also, the base should be the same while adding or subtracting the powers.