log-3-a-1-log-4-a-8-a-please-solution-

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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