+0  
 
0
30
3
avatar+702 

10^(z+2)=27

critical  Nov 23, 2018
 #1
avatar+152 
+2

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  Nov 23, 2018
 #2
avatar+92429 
+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

 

 

cool cool cool

CPhill  Nov 23, 2018
 #3
avatar+152 
+1

That seems better! Thank you, CPhill! 

azsun  Nov 23, 2018

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