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Question Number 104199 by hardylanes last updated on 20/Jul/20

s o l v e f o r r e a l v a l u e s o f x t h e e q u a t i o n

4 (3^{2 x + 1}) + 17 (3^{x}) = 7 .

i f m a n d n a r e p o s i t i v e r e a l n u m b e r s o t h e r

t h a n 1, s h o w t h a t t h e \log_{n} m + \log_{\frac{1}{m}} n = 0

Answered by bemath last updated on 20/Jul/20

12 . 3^{x} + 17 . 3^{x} - 7 = 0

3^{x} = \frac{- 17 + \sqrt{289 + 48.7}}{24}

3^{x} = \frac{- 17 + 25}{24} = \frac{1}{3}

∴ x = - 1

Answered by OlafThorendsen last updated on 20/Jul/20

4 (3^{2 x + 1}) + 17 (3^{x}) = 7

12 {(3^{x})}^{2} + 17 (3^{x}) - 7 = 0

u = 3^{x}

12 u^{2} + 17 u - 7 = 0

Δ = 17^{2} - 4 \times 12 \times (- 7) = 625 = 25^{2}

u = \frac{- 17 \pm 25}{24}

u = 3^{x} \Rightarrow u > 0

then u = \frac{- 17 + 25}{24} = \frac{1}{3}

u = 3^{x} = \frac{1}{3} \Rightarrow x = - 1

I think your second

question is false

({it}^{'} s true only for m = n

or m = \frac{1}{n}) but :

\log_{n} m + \log_{\frac{1}{n}} m =

\frac{\ln m}{\ln n} + \frac{\ln m}{\ln \frac{1}{n}} =

\frac{\ln m}{\ln n} - \frac{\ln m}{\ln n} = 0