Question

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Solve this attachment question

NOTE >Solve both questions 15 and 16
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Answer 1

Answer:

thank you for see the answer

Answer 2

15.

$\frac{1}{1 + {x}^{a - b} } + \frac{1}{1 + {x}^{b - a} }$

LCM=Product of Denominator.

LCM=[1+x^(a-b)]×[1+x^(b-a)]

$\frac{1(1 + {x}^{b - a}) +1(1 + {x}^{a - b} ) }{(1 + {x}^{a - b})( 1 + {x}^{b - a} )} \\ \\ \frac{1 + {x}^{b - a} + 1 + {x}^{a- b} }{1 + {x}^{b - a} + {x}^{a - b} + {x}^{a - b} \times {x}^{b - a} } \\ \\ \frac{2 + {x}^{b - a} + {x}^{a - b} }{1 + {x}^{b - a} + {x}^{a - b} + {x}^{a - b + b - a} } \\ \\ \frac{2 + {x}^{b - a} + {x}^{a - b} }{1 + {x}^{b - a} + {x}^{a - b} + {x}^{0} } \\ \\ \frac{2 + {x}^{b - a} + {x}^{a - b} }{1 + {x}^{b - a} + {x}^{a - b} + 1 } \\ \\ \cancel{ \frac{2 + {x}^{b - a} + {x}^{a - b} }{2+ {x}^{b - a} + {x}^{a - b} }} \\ \\ = 1$

16.

${25}^{x - 1} = {5}^{2x - 1} - 100 \\ \\ {5}^{2(x - 1)} = {5}^{2x - 1} - 100 \\ \\ {5}^{2x - 2} = {5}^{2x - 1} - 100 \\ \\ {5}^{2x} \times {5}^{ - 2} = {5}^{2x} \times {5}^{ - 1} - 100 \\ \\ {5}^{2x} \times \frac{1}{ {5}^{2} } = {5}^{2x} \times \frac{1}{5} - 100 \\ \\ \frac{ {5}^{2x} }{25} = \frac{ {5}^{2x} }{5} - 100 \\ \\ \frac{ {5}^{2x} }{25} - \frac{ {5}^{2x} }{5} = - 100 \\ \\ {5}^{2x} ( \frac{1}{25} - \frac{1}{5} ) = - 100 \\ \\ {5}^{2x} \times ( \frac{1 - 1 \times 5}{25} ) = - 100 \\ \\ {5}^{2x} \times \frac{ - 4}{25} = - 100 \\ \\ {5}^{2x } = - 100 \times \frac{ 25}{ - 4} \\ \\ {5}^{2x} = 25 \times 25 \\ \\ {5}^{2x} = {5}^{2} \times {5}^{2} \\ \\ {5}^{2x} = {5}^{2 + 2} \\ \\ {5}^{2x} = {5}^{4} \\ \\ 2x = 4 \\ \\ x = \frac{4}{2} \\ \\ x = 2$

REMEMBER

${a}^{x} \times {a}^{y} = {a}^{x + y} \\ \\ {a}^{0} = 1 \\ \\ {a}^{ - 1} = \frac{1}{a}$