we may use this logarithmic identity by EULER, log (base b) a = (log a)/(log b)
solve for x: log3(x+1)=log9(26+2x)
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log (x + 1)/(log 3) = log (2x + 26)/(log 9) = log (2x + 26)/(2 log 3)
cross-multiply,
2 (log 3) log (x + 1) = (log 3) log (2x + 26), cancel log 3
2 log (x + 1) = log (2x + 26),
log (x + 1)^2 = log (2x + 26),
take anti-log og both sides,
(x + 1)^2 = 2x + 26,
x^2 + 2x + 1 = 2x + 26,
2x^2 - 2x + 2x - 26 +1 = 0,
x^2 - 25 = 0
factoring,
(x + 5)(x - 5) =0,
x = {-5, 5}, but only 5 is the valid answer
the graph below is wrong