solve for real values of x the equation 4(3^(2x+1) )+17(3^x )=7. if m and n are positive real numbers other than 1, show that the log_n m+log


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
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

	$4 (3^{2 x + 1}) + 17 (3^{x}) = 7$ $12 {(3^{x})}^{2} + 17 (3^{x}) - 7 = 0$ $u = 3^{x}$ $12 u^{2} + 17 u - 7 = 0$ $Δ = 17^{2} - 4 \times 12 \times (- 7) = 625 = 25^{2}$ $u = \frac{- 17 \pm 25}{24}$ $u = 3^{x} \Rightarrow u > 0$ $then u = \frac{- 17 + 25}{24} = \frac{1}{3}$ $u = 3^{x} = \frac{1}{3} \Rightarrow x = - 1$ $I think your second$ $question is false$ $({it}^{'} s true only for m = n$ $or m = \frac{1}{n}) but :$ $\log_{n} m + \log_{\frac{1}{n}} m =$ $\frac{\ln m}{\ln n} + \frac{\ln m}{\ln \frac{1}{n}} =$ $\frac{\ln m}{\ln n} - \frac{\ln m}{\ln n} = 0$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum