Find-the-value-of-x-9-x-6-x-4-x-Please-help-

Question Number 5685 by sanusihammed last updated on 24/May/16

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

9^{x} = 6^{x} + 4^{x}

P l e a s e h e l p .

Answered by prakash jain last updated on 24/May/16

Divide by 4^{x}

{(\frac{9}{4})}^{x} = {(\frac{6}{4})}^{x} + 1

{(\frac{3}{2})}^{2 x} - {(\frac{3}{2})}^{x} - 1 = 0

{(\frac{3}{2})}^{x} = u

u^{2} - u - 1 = 0

\Rightarrow u = \frac{1 \pm \sqrt{5}}{2}

for real x, u > 0 \Rightarrow u = \frac{1 + \sqrt{5}}{2}

{(\frac{3}{2})}^{x} = \frac{1 + \sqrt{5}}{2}

x (\ln 3 - \ln 2) = \ln (1 + \sqrt{5}) - \ln 2

x = \frac{\ln (1 + \sqrt{5}) - \ln 2}{\ln 3 - \ln 2}

Commented by sanusihammed last updated on 24/May/16

T h a n k y o u f o r h o u r h e l p .

Commented by Rasheed Soomro last updated on 24/May/16

N^{i} CE!