Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# Find the value x, if ${{5}^{2x-1}}-{{25}^{x-1}}=2500$ .

Last updated date: 18th Jun 2024
Total views: 393.6k
Views today: 6.93k
Verified
393.6k+ views
Hint: Convert the terms involving the variable x, in powers of the same number.
It can be observed that $25={{5}^{2}}$ .
The following properties of indices (powers) are useful.
(i) ${{a}^{x}}\times {{a}^{y}}={{a}^{x+y}}$
(ii) $\dfrac{{{a}^{x}}}{{{a}^{y}}}={{a}^{x-y}}$
(iii) ${{({{a}^{x}})}^{y}}={{a}^{xy}}$

Let us start simplifying by converting all the terms which have the variable x, into powers of 5.
${{5}^{2x-1}}-{{25}^{x-1}}=2500$
⇒ ${{5}^{2x-1}}-{{({{5}^{2}})}^{x-1}}=2500$
Using the property that ${{({{a}^{x}})}^{y}}={{a}^{xy}}$ , the above expression equals:
⇒ ${{5}^{2x-1}}-{{5}^{2}}^{x-2}=2500$
And, using the property that $\dfrac{{{a}^{x}}}{{{a}^{y}}}={{a}^{x-y}}$ , the above can be written as:
⇒ $\dfrac{{{5}^{2x}}}{{{5}^{1}}}-\dfrac{{{5}^{2x}}}{{{5}^{2}}}=2500$
Let us substitute ${{5}^{2x}}=y$ to reduce writing work and a simpler look of the equation:
⇒ $\dfrac{y}{5}-\dfrac{y}{25}=2500$
Multiply both the numerator and denominator of $\dfrac{y}{5}$ by 5, to get same denominators on the LHS:
⇒ $\dfrac{5y}{25}-\dfrac{y}{25}=2500$
Multiplying both sides of the equation by 25, so that we are left with only the variable on the LHS:
⇒ 5y - y = 25 × 2500
⇒ 4 y = 25 × 25 × 100
Dividing both sides of the equation by 4:
⇒ y = 25 × 25 × 25
⇒ $y={{5}^{2}}\times {{5}^{2}}\times {{5}^{2}}$
Using the property ${{a}^{x}}\times {{a}^{y}}={{a}^{x+y}}$ , we get:
⇒ $y={{5}^{2+2+2}}$
⇒ $y={{5}^{6}}$ .
In order to solve for x, let us back-substitute ${{5}^{2x}}=y$ in the above value of y.
⇒ ${{5}^{2x}}={{5}^{6}}$
Since the bases are the same, the powers must also be the same.
∴ 2x = 6
⇒ x = 3.
Hence, the value of x is 3.

Note: If ${{a}^{x}}{{b}^{y}}={{a}^{p}}{{b}^{q}}$ , then it is not necessarily true that x = p and y = q.
But if ${{a}^{x}}={{a}^{p}}$ , then it is always true that x = p.