Question

please solve the problem.

Answer 1

$Given: 25^{x - 1} = 5^{2x - 1} - 100$

$=> 25^{x - 1} - 5^{2x - 1} = -100$

$=> (5^2)^{x - 1} - 5^{2x - 1} = -100$

$=> 5^{2x - 2} - 5^{2x - 1} = -100$

$=> 5^{2x} * 5^{-2} - 5^{2x} . 5^{-1} = -100$

$=> 5^{2x}(5^{-2} - 5^{-1}) = -100$

$=> 5^{2x} (\frac{1}{5^2} - \frac{1}{5}) = -100$

$=> 5^{2x}(\frac{1}{25} - \frac{1}{5}) = -100$

$=> 5^{2x}(\frac{-4}{25}) = -100$

$=> 5^{2x} = \frac{-100}{\frac{-4}{25} }$

$=> 5^{2x} = \frac{2500}{4}$

$=> 5^{2x} = 625$

$=> 5^{2x} = 5^4$

$=> 2x = 4$

$=> \boxed{x = 2}$

Hope this helps!

Answer 2

Hi there !

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Given :

$25 {}^{x - 1} = 5 {}^{2x - 1} - 100$

Solution :

$25 {}^{x - 1} - 5 {}^{2x - 1} = - 100 \\ \\( 5 {}^{2} ) {}^{x - 1} - 5{}^{2x - 1} = - 100 \\ \\ 5{}^{2x - 2} - 5{}^{2x - 1} = - 100 \\ \\ 5 {}^{2x} \times 5 {}^{ - 2} - 5^{2x} \times 5 {}^{ - 1} = 100 \\ \\ 5^{2x}( 5 {}^{ - 2} - 5 {}^{ - 1} ) = - 100 \\ \\ 5^{2x} ( \frac{1}{5 {}^{2} } - \frac{1}{5} ) = - 100 \\ \\ 5^{2x} ( \frac{1}{25} - \frac{1}{5} ) = - 100 \\ \\ 5^{2x} ( \frac{1 - 5}{25} ) = - 100 \\ \\ 5^{2x} ( \frac{ - 4}{25} ) = - 100 \\ \\ 5^{2x} = \frac{ \cancel- 100}{ \frac{ \cancel - 4}{25} } \\ \\ 5^{2x} = \frac{ \cancel{2500} \: {}^{625} }{ \cancel{ \:4 \: \: _1 } } \\ \\ 5 {}^{2x} = 625 \\ \\ \cancel 5 \: {}^{2x} = \cancel 5 \: {}^{4} \\ \\ 2x = 4 \\ \\ x = \frac{ \cancel {4} \: {}^{2} }{ \cancel 2} \\ \\ \therefore \boxed{\bold{x = 2}}$
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Thanks for the question !