What is the solution of the equation log15 [1/(3^x + x - 5)]=x*(log15 5 - 1) ?

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We have to solve log(15) [1/(3^x + x - 5)] = x * (log(15) 5 - 1).

log(15) [1/(3^x + x - 5)] = x * (log(15) 5 - 1)

=> log(15) [1/(3^x + x - 5)] = x*log(15) 5 - x

=> log(15) [1/(3^x + x - 5)] =...

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We have to solve log(15) [1/(3^x + x - 5)] = x * (log(15) 5 - 1).

log(15) [1/(3^x + x - 5)] = x * (log(15) 5 - 1)

=> log(15) [1/(3^x + x - 5)] = x*log(15) 5 - x

=> log(15) [1/(3^x + x - 5)] = x*log(15) 5 - log (15) 15^x

=> log(15) [1/(3^x + x - 5)] = log(15) (5^x/ 15^x)

=> [1/(3^x + x - 5)] = 5^x/ 15^x

=> [1/(3^x + x - 5)] = 1/ 3^x

=> 3^x = 3^x + x - 5

=> x - 5 = 0

=> x = 5

We get x = 5

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