log-2x-1-x-2-1-log-x-1-3-2x-5-18-2x-

Question Number 143329 by Huy last updated on 13/Jun/21

\log_{2 x + 1} (x^{2} + 1) + \log_{x + 1} (3 \sqrt{2 x + 5} + 18) = 2 x

Answered by Olaf_Thorendsen last updated on 13/Jun/21

\log_{2 x + 1} (x^{2} + 1) + \log_{x + 1} (3 \sqrt{2 x + 5} + 18) = 2 x

If 2 x + 1 = x^{2} + 1 (x = 2) :

\log_{2 x + 1} (x^{2} + 1) = 1

\log_{x + 1} (3 \sqrt{2 x + 5} + 18) = \log_{3} 27 = 3

2 x = 2 \times 2 = 4

x = 2 is a trivial solution