Logs???

daniella · January 17th 2007, 05:51 PM

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!

**ThePerfectHacker** · January 17th 2007, 06:18 PM

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

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

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

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

**Soroban** · January 17th 2007, 09:06 PM

Hello, daniella!

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

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

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

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

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

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

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

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

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

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

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

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