Question

Can anyone solve this sum. plz don't spam ​

Answer 1

Required Answer:-

Given question:

$\large{ \boxed{\rm{ {5}^{2x - 1} = \frac{1}{ {125}^{x - 3} } \: \: find \: x }}}$

Here,

• First express 125 in terms of power of 5.
• Nextly, equalise the bases i.e. 5
• Compute the relation between powers of 5 in both sides of the equation.

Solving...

Expressing in terms of 5 both sides,

➛$\rm{5}^{2x - 1} = \dfrac{1}{ ({5}^{3}) {}^{x - 3} }$

When the power is negative, it means the value is in the denominator with the same magnitude power (exponent).

➛$\rm{ {5}^{2x - 1} = \dfrac{1}{ {5}^{3x - 9} } }$

➛$\rm{5}^{2x - 1} = {5}^{ - (3x - 9)}$

Now, the bases are same. We can directly do the computation upon the powers.

➛$\rm2x - 1 = - (3x - 9)$

➛$\rm{2x - 1 = - 3x + 9}$

➛$\rm{2x + 3x = 9 + 1}$

➛$\rm 5x = 10$

➛$\rm{\red{x = 2}}$

Hence:-

• The required value of x, as asked in the question is 2 (Ans).

Answer 2

Given :-

$\large{\bf{{5}^{2x - 1} = \frac{1}{ {125}^{x - 3} } \: find \: x}}$

To find :-

• First expression 125 in terms of power of 5.
• Nextly , equalise the base i.e, 5
• Compute the relation between powers of 5 in both sides of the equation.

Solution :-

Expressing in terms of 5 both sides ,

$\large\implies$ $\large{\tt{{5}^{2x - 1} = \frac{1}{ {(5³)}^{x - 3} }}}$

When the power is negative , It means the value is in the denominator with the same magnitude power (exponent).

$\large\implies$ $\large{\tt{ {5}^{2x - 1} = \frac{1}{ {5}^{3x - 9} } }}$

$\large\implies$ $\large{\tt{{5}^{2x - 1} = {5}^{ - (3x - 9)} }}$

Now,

The bases are same. We can directly do the computation upon the powers.

$\large\implies$ $\large{\tt{2x - 1 = - (3x - 9)}}$

$\large\implies$ $\large{\tt{2x - 1 = - 3x + 9}}$

$\large\implies$ $\large{\tt{2x + 3x = 9 + 1}}$

$\large\implies$ $\large{\tt{5x = 10}}$

$\large\implies$ $\boxed{\bf\green{x = 2}}$

Hence,

• The required value of x , as asked in the question in 2 (Ans). $\large{\bf\green{✓}}$