# Math Help - another logarithm problem

1. ## another logarithm problem

can someone help me with this one?

4^3p*4^2p-1=100 the answer is .8644 thanks!!

i need someone to show me to break it down step by step.

i did 4^3p*4^2p-1 =4^5p-1 is that right or would it be=16^5p-1?

2. ## Logs

Hello Evandmal07
Originally Posted by Evandmal07
can someone help me with this one?

4^3p*4^2p-1=100 the answer is .8644 thanks!!

i need someone to show me to break it down step by step.

i did 4^3p*4^2p-1 =4^5p-1 is that right or would it be=16^5p-1?
You were right first time: $4^{3p} \times 4^{2p-1} = 4^{5p-1}$.

The key to solving this type of problem is the log formula:

• $\log(a^b)=b\log (a)$

So if you now take logs of both sides (base 10 is easiest):

$\log(4^{5p-1})=\log(100) = \log(10^2) = 2\log(10) =2$

$\Rightarrow (5p-1)\log 4= 2$

$\Rightarrow5p-1 =\frac{2}{\log 4}$

$\Rightarrow p =\frac15\Big(\frac{2}{\log 4}+1\Big)=0.8644$ (to 4 d.p.)

3. ## thank you!!!

thank you so much now i see!!! i just didnt realize that i needed to convert that log100 to 2 but i see that that makes sense!!

wow thanks so much!!!