How do you solve log2sqrt(5x+1) = 4log2√5x+1=4?

1 Answer
Gió
Sep 11, 2015

I found:
x=5xx10^6x=5×106 if log in base 10
or
x=51x=51 if log in base 2

Explanation:

Assuming a log in base 1010 we can write:
log_(10)2sqrt(5x+1)=4log102√5x+1=4

using the definition of log:
2sqrt(5x+1)=10^42√5x+1=104
cancel(2)sqrt(5x+1)=10000/2
sqrt(5x+1)=5000

squaring both sides:
5x+1=(5000)^2
x=((5000)^2-1)/5~~5xx10^6

If on the other hand the base is 2 we can write:
log_2sqrt(5x+1)=4
again using the definition of log we get:
sqrt(5x+1)=2^4
sqrt(5x+1)=16
squaring both sides:
5x+1=256
5x=255
x=51

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