The value of x satisfying
`x+log_10(1+2^x)=x log_10 5+log_10 6` is

Explanation

`x+log_10(2^x+1)=xlog_10 5+log_10 6`

`log_10 10^x=xlog_10 10=lambda`

`log_10 10^lambda+log_10(2^lambda+1)=log_10 5^x+log_10 6`

`2^x*5^x[2^x+1]=6*5^x=5^lambda[2^(2lambda(+2^lambda-6))]=0`

`5^x=0,2^(2lambda)+2^lambda-6=0`

`2^x=t^2+t-6=0`

`t^2+3t-2t-6=0`

`t(t+30-2(t+3)=0`

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