5-x-5-x-7-100-Find-the-value-of-x-

Question Number 6483 by sanusihammed last updated on 28/Jun/16

5^{\sqrt{x}} - 5^{(x - 7)} = 100

F i n d t h e v a l u e o f x

Answered by Yozzii last updated on 29/Jun/16

5^{\sqrt{x}} - 5^{- 7} 5^{x} = 100

L e t u = \sqrt{x} \Rightarrow u^{2} = x

∴ 5^{u} - 5^{- 7} 5^{u^{2}} = 100

5^{u} - 5^{u^{2} - 7} = 100 = 2^{2} 5^{2}

F o r b o t h l . h . s a n d r . h . s t o b e i n t e g e r s (100)

t h e n u^{2} - 7 ⩾ 0 \Rightarrow u^{2} ⩾ 7 . T r y u = 3 i f u \in Z .

\Rightarrow u^{2} ⩾ 7 a n d 5^{3} - 5^{9 - 7} = 125 - 25 = 100 .

S o, x = u^{2} = 9 s a t i s f i e s t h e g i v e n e q u a t i o n .

T h e r e p o s s i b l y e x i s t s o t h e r v a l u e s o f

x s i n c e t h i s d o e s n o t p r o v i d e p r o o f o f

x = 9 a s t h e o n l y a n s w e r .

I f l . h . s > 0, a s 100 > 0, w e r e q u i r e t h a t

u > u^{2} - 7

u^{2} - u - 7 < 0

(u - \frac{1 + \sqrt{29}}{2}) (u - \frac{1 - \sqrt{29}}{2}) < 0

\Rightarrow \frac{1 - \sqrt{29}}{2} < u < \frac{1 + \sqrt{29}}{2}

\frac{1 - \sqrt{29}}{2} < \sqrt{x} < \frac{1 + \sqrt{29}}{2}

{(\frac{1 - \sqrt{29}}{2})}^{2} < x < {(\frac{1 + \sqrt{29}}{2})}^{2}

4.807 \dots < x < 10.19 \dots

5^{\sqrt{x}} - 5^{x - 7} = 100 .

5^{\sqrt{x} + 7 - x} - 1 = 2^{2} 5^{9 - x}

Commented by sanusihammed last updated on 29/Jun/16

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