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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:
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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