log_3 (a+1)=log_4 (a+8) a=?? please solution


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Question Number 185085 by Ml last updated on 16/Jan/23

$\log_{3} (a + 1) = \log_{4} (a + 8)$ $a = ? ?$ $please solution$
	Answered by Frix last updated on 16/Jan/23

	$\log_{3} (a + 1) = x \Leftrightarrow 3^{x} = a + 1 \Leftrightarrow a = 3^{x} - 1$ $\log_{4} (a + 8) = x \Leftrightarrow 4^{x} = a + 8 \Leftrightarrow a = 4^{x} - 8$ $\Rightarrow$ $4^{x} - 8 = 3^{x} - 1$ $4^{x} - 3^{x} = 7$ $Obviously x = 2 \Rightarrow$ $a = 3^{2} - 1 = 8$ $a = 4^{2} - 8 = 8$
	Answered by aba last updated on 16/Jan/23

	$consider the function$ $f (x) = \log_{3} (x + 1) - \log_{4} (x + 8)$ $Df =] - 1, + \infty [$ $f^{^{'}} (x) = \frac{1}{(x + 1) \ln (3)} - \frac{1}{(x + 8) \ln (4)}$ $= \frac{(x + 8) \ln (4) - (x + 1) \ln (3)}{(x + 1) (x + 8) \ln (3) \ln (4)}$ $= \frac{x (\ln (4) - \ln (3)) + 8 \ln (4) - \ln (3)}{(x + 1) (x + 8) \ln (3) \ln (4)}$ $the denominator is positive on the domain of f$ $f^{'} (x) = 0 \Rightarrow x (\ln (4) - \ln (3)) + 8 \ln (4) - \ln (3) = 0$ $\Rightarrow x = - \frac{8 \ln (4) - \ln (3)}{\ln (4) - \ln (3)}$ $\Rightarrow x = - \frac{\ln (\frac{4^{8}}{3})}{\ln (\frac{4}{3})} < - 1$ $\Rightarrow x < - 1$ $so the function f is strictly increasing on] - 1, + \infty [. Hence the equation f (x) = 0 has at most one solution$ $since f (8) = 0, we found it$
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