How do you solve #5^(x+2) = 2#?

2 Answers
Nov 10, 2015

#x=(log(2/25))/(log5)=-1,57#

Explanation:

From the laws of exponents we may write this as

#5^x*5^2=2#

Now dividing throughout by #5^2# yields

#5^x=2/25#

Now taking the log on both sides and using laws of logs we can write it as

#xlog5=log(2/25)#

Finally dividing throughout by #log5# gives

#x=(log(2/25))/(log5)=-1,57#

Nov 14, 2015

To undo the #x# being raised to an exponent, take the log base 5 of both sides. This undoes the 5 on the left, leaving just #x+2#. On the right is #log_"5"2#, which can be put into a calculator using the change of base formula, in which #log_"5"2=(log2)/(log5)#.

Then, 2 is subtracted from both sides, leaving #x# equal to #(log2)/(log5)-2~~-1.57#.