log-5-x-3-log-6-x-14-

Question Number 170958 by cortano1 last updated on 04/Jun/22

\log_{5} (x + 3) = \log_{6} (x + 14)

Answered by floor(10²Eta[1]) last updated on 05/Jun/22

\log_{5} (x + 3) = y = \log_{6} (x + 14)

x + 3 = 5^{y}

x + 14 = 6^{y}

5^{y} + 11 = 6^{y}

y = 2 \Rightarrow 5^{2} + 11 = 6^{2}

{let}^{'} s show that for y > 2 \Rightarrow 6^{y} > 5^{y} + 11

let f (y) = 6^{y} - 5^{y} - 11

we want : f (y) > 0, \forall y > 2

f (2) = 0

f^{'} (y) = 6^{y} \ln 6 - 5^{y} \ln 5

but f^{'} (y) > 0, \forall y > 2

\Rightarrow f is increasing on the interval [2, + \infty)

since f (2) = 0 \Rightarrow f (y) > 0, \forall y > 2

\Rightarrow 6^{y} > 5^{y} + 11, \forall y > 2

\Rightarrow y = 2 is the only solution to 6^{y} = 5^{y} + 11

\Rightarrow x + 3 = 5^{2} \Rightarrow x = 22