# How solve x=?y=? 10^(1+log(3x-1))=log(y+1) x+log(y^2+2y+1)=71/30i have no idea how began

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### 1 Answer

You need to use the rules of exponentiation such that:

`10^(1+log(3x-1)) = 10^1*10^(log(3x-1))`

You should use the logarithmic identity `a^(log_a b) = b` such that:

`10^(log(3x-1)) = 3x - 1`

`10^(1+log(3x-1)) = 10*(3x-1)`

Notice that you may rewrite the first equation such that:

`10*(3x-1) = log(y+1)`

Considering the logarithm in the second equation yields:

`log(y^2+2y+1)=log(y+1)^2`

Using the logarithmic identity `log x^n = n*log x` such that:

`log(y^2+2y+1) = 2log(y+1)`

You may rewrite the second equation such that:

`x + 2log(y+1) = 71/30`

Notice that you may substitute `10*(3x-1)` for `log(y+1)` such that:

`x + 20(3x-1) = 71/30`

`30x + 600(3x-1) = 71 => 30x + 1800x - 600 = 71`

`1830x = 71 => x = 71/1830 ~~ 0.03`

Since x needs to be larger than `1/3~~ 0.3` , then the value of x is invalid.

Since the value of x is invalid, hence the value of y cannot be evaluated.

**Hence, evaluating the solution to the given equation yields that there are no solutions.**

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