Question

how do you solve 4^x=20?

Answer 1

well, I'd use the aforementioned logarith cancellation rule :)

Answer 2

\[a^x=b\]
\[x=\log _ab\]

Answer 3

thanks you alot guys i apreciate it (:

Answer 4

\(\bf \text{now using => }log_aa^x = x\\
4^x = 20\\
log_44^x = log_420 \implies x = log_420\)

Answer 5

then you'd use the "change of base rule" to get the value using log base 10

Answer 6

that means i have to divide right?

Answer 7

log change of base rule => \(\bf log_ab = \cfrac{log_{10}b}{log_{10}a}\)

Answer 8

is it 2.16?

Answer 9

that's what I got, yes

Answer 10

why not just use base e or base 10 logs\

\[x = \frac{\ln(20)}{\ln(4)}\]

don't worry about change of base

Answer 11

or you could just take logs to base e of both sides

ln 4^x = ln 20
and by the law of logs:
x ln4 = ln 20

x = lm20 / ln 4