Thread: Logs???

how do i use the "change of base" formula to solve

1 a) 3^2x=8
it is the exponent 2x
2 d) 5^x-2=150
it is to the exponent x-2

Thank You!

Originally Posted by daniella

how do i use the "change of base" formula to solve

1 a) 3^2x=8
it is the exponent 2x

$\displaystyle 3^{2x}=8=2^3$
Thus,
$\displaystyle 2x=3$
$\displaystyle x=1.5$

2 d) 5^x-2=150
it is to the exponent x-2

$\displaystyle 5^{x-2}=150$
Thus,
$\displaystyle \log_5 150=x-2$
$\displaystyle x=2+\log_5 150$

Hello, daniella!

You're expected to know that $\displaystyle b^x\:=\:a$ is equivalent to $\displaystyle x \:=\:\log_ba$
. . and the Base-change formula: .$\displaystyle \log_aX \;=\;\frac{\log_bX}{\log_ba}$

How do i use the "change of base" formula to solve:

$\displaystyle 1\:a)\;3^{2x}\:=\:8$

$\displaystyle 2\:d)\;5^{x-2}\:=\:150$

"Change of Base" formula: .$\displaystyle \log_aX \:=\:\frac{\log_bX}{\log_ba}$


1a) We have: .$\displaystyle 3^{2x} \:=\:8$ . . . which is equivalent to: .$\displaystyle 2x \:=\:\log_38$

Change-of-base: .$\displaystyle 2x\:=\:\frac{\log8}{\log3} $

Therefore: .$\displaystyle x \;=\;\frac{\log8}{2\cdot\log3} \;\approx\;0.9464$


2d) We have: .$\displaystyle 5^{x-2}\;=\;150$ . . . which is equivalent to: .$\displaystyle x-2 \;=\;\log_5150$

Change-of-base: .$\displaystyle x-2\;=\;\frac{\log150}{\log5}$

Therefore: .$\displaystyle x \;=\;\frac{\log150}{\log5} + 2 \;\approx\;5.1133$