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