+199 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 \(z=\frac{3\ln \left(3\right)}{\ln \left(10\right)}-2.\)
+129852 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
