Question

If \[{5^{2x - 1}} - {\left( {25} \right)^{x - 1}} = 2500\] ,then the value of x is
A.2
B.5
C.3
D.1

Answer 1

Hint: We can write 25 as a square of 5. Then we will use the laws of indices and find the value of x. A negative exponent means how many times to divide one by the number.

Complete step-by-step answer:
 Given that
\[{5^{2x - 1}} - {\left( {25} \right)^{x - 1}} = 2500\]
25 can be written as \[{\left( 5 \right)^2}\].
\[ \Rightarrow {5^{2x - 1}} - {\left( {{{\left( 5 \right)}^2}} \right)^{x - 1}} = 2500\]
Using law of indices,
\[ \Rightarrow {5^{2x - 1}} - {\left( 5 \right)^{2x - 2}} = 2500...... \to {\left( {{a^m}} \right)^n} = {a^{mn}}\]
Again,
\[ \Rightarrow {5^{2x}}{5^{ - 1}} - {5^{2x}}{5^{ - 2}} = 2500...... \to {a^{m + n}} = {a^m}{a^n}\]
Taking \[{5^{2x}}\] common
\[ \Rightarrow {5^{2x}}({5^{ - 1}} - {5^{ - 2}}) = 2500\]
\[ \Rightarrow {5^{2x}}\left( {\dfrac{1}{5} - \dfrac{1}{{{5^2}}}} \right) = 2500\]
Writing the power of 5,
\[ \Rightarrow {5^{2x}}\left( {\dfrac{1}{5} - \dfrac{1}{{25}}} \right) = 2500\]
Taking LCM of the terms of bracket as 25
\[\begin{gathered}
   \Rightarrow {5^{2x}}\left( {\dfrac{{5 - 1}}{{25}}} \right) = 2500 \\
   \Rightarrow {5^{2x}} \times \dfrac{4}{{25}} = 2500 \\
\end{gathered} \]
Rearranging the terms
\[ \Rightarrow {5^{2x}} = \dfrac{{2500 \times 25}}{4}\]
Cancelling 2500 by 4
\[ \Rightarrow {5^{2x}} = 625 \times 25\]
Now write the numbers on RHS in power of 5
\[ \Rightarrow {5^{2x}} = {5^4} \times {5^2}\]
Again we will use laws of indices
\[ \Rightarrow {5^{2x}} = {5^{4 + 2}}...... \to {a^{m + n}} = {a^m}{a^n}\]
\[ \Rightarrow {5^{2x}} = {5^6}\]
Now since base are same on LHS and RHS so just compare the powers
\[ \Rightarrow 2x = 6\]
Divide 6 by 2
\[ \Rightarrow x = \dfrac{6}{2}\]
There we find our answer
\[ \Rightarrow x = 3\]
Thus the value of x is 3.
Hence the correct option is C.

Note: In the problem above, we should simply use the laws of indices. Students might get confused in using the laws correctly in the correct place. So for this don’t skip steps .Write even a single step.
\[ \to {\left( {{a^m}} \right)^n} = {a^{mn}}\]
\[ \to {a^{m + n}} = {a^m}{a^n}\].
We can check options also to get the answer.