SOLUTION: Please show me how to solve for x here: 4^2x = 16
Appreciate your help. Thanks!
Algebra.Com
Question 1093270: Please show me how to solve for x here: 4^2x = 16
Appreciate your help. Thanks!
Found 2 solutions by Theo, Alan3354:
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
4^(2x) = 16
take the log of both sides to get log(4^(2x)) = log(16)
this is equivalent to 2x * log(4) = log(16)
divide both sides by log(4) to get 2c = log(16)/log(4)
divide both sides by 2 to get x = log(16)/log(4)/2
use your calculator to get x = 1
you could also have done:
4^(2x) = 16
since you know that 4^2 = 16, then you get:
4^(2x) = 16 and 4^2 = 16
since they're both equal to 16 then they're both equal to each other and you get:
4^2 = 4^(2x)
this can only be true if 2 = 2x.
solve for x to get x = 1.
note that the properties of logs and exponents that were used were:
log(x^a) = a*log(x)
if x^a = x^b, then a = b
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
16x = 16
x = 1
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