10^(z+2)=27

Translating this, we get $\ln \left(10^{z+2}\right)=\ln \left(27\right).$ Now, we just apply the log rule($\log _a\left(x^b\right)=b\cdot \log _a\left(x\right)$) , and get  $\left(z+2\right)\ln \left(10\right)=\ln \left(27\right).$

Then, we can simplify this to be  $\left(z+2\right)\ln \left(10\right)=3\ln \left(3\right).$So, we solve it, and our final answer is $z=\frac{3\ln \left(3\right)}{\ln \left(10\right)}-2.$

Azsun's answer is perfectly valid...but..because we have 10 raised to a power..using the base 10 log seems more natural

log 10^(z + 2)   = log 27      and we can write 

(z + 2)  log 10   = log 27          [log 10 = 1...so...we can ignore this ]

z + 2    =  log 27        subtract 2 from both sides

z = log 27 - 2  ≈  -0.569