1) Log(4)64=x 2) Log(3)243=2x+1 The numbers in parenthesis is the base. I know the first one I can just throw in the calculator, but that's now how we're supposed to solve these ones. What's the value of x?

1 Answer
Gió
Oct 10, 2015

Have a look:

Explanation:

In the first you do not need the calculator.
Consider:
#log_4(64)=x#
use the definition of log...what is the number #x# you need to raise #x# to get #64#?
or: #4^x=64#
you can see that is #3# in fact #4^3=64#
so #x=3#
Also the second do not require the calculator in fact you know that #3^5=243# so you can write:
#log_3(3^5)=2x+1# use again the definition of log:
#3^5=3^(2x+1)# so that:
#5=2x+1#
#x=4/2=2#

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