# [SOLVED] Reviewing Logs and stuck on a problem...help?!?

• Oct 6th 2009, 10:40 PM
strngFan
[SOLVED] Reviewing Logs and stuck on a problem...help?!?
I have the following problem on my homework:

50(1.02)^3x = 30(1.47)^x solve for x

I've tried many things and quick math gave the answer as x=1.56... but I am not able to reproduce this any way I have considered. What am I doing wrong?

Here was how I have approached this problem most recently:
30/50 = (1.02)^3x / (1.47)^x

Is it necessary to use change of base on the RHS top and bottom and then substitue back in? I ask because that is in the section which follows the one we are covering....

• Oct 6th 2009, 11:08 PM
Hello strngFan

Welcome to Math Help Forum!
Quote:

Originally Posted by strngFan
I have the following problem on my homework:

50(1.02)^3x = 30(1.47)^x solve for x

I've tried many things and quick math gave the answer as x=1.56... but I am not able to reproduce this any way I have considered. What am I doing wrong?

Here was how I have approached this problem most recently:
30/50 = (1.02)^3x / (1.47)^x

Is it necessary to use change of base on the RHS top and bottom and then substitue back in? I ask because that is in the section which follows the one we are covering....

Thanks for showing us your working, but the heading tells you what to do: take logs of both sides.

$50(1.02)^{3x} = 30(1.47)^x$

$\Rightarrow \log(50(1.02)^{3x}) = \log(30(1.47)^x)$

$\Rightarrow \log50+3x\log1.02 = \log30+x\log1.47$ (Do you understand this step? It's just using the Laws of Logarithms.)

Now solve for $x$ in the usual way. Here's the next step:

$3x\log1.02-x\log1.47=\log30-\log50$

Can you complete it now?

• Oct 6th 2009, 11:11 PM
strngFan
thank you very much!
Thank you so much for your help! Thought it had to be simpler than what I was making it into. Have a great night!!

Sarah
• Oct 6th 2009, 11:50 PM
earboth
Quote:

Originally Posted by strngFan
I have the following problem on my homework:

50(1.02)^3x = 30(1.47)^x solve for x

I've tried many things and quick math gave the answer as x=1.56... but I am not able to reproduce this any way I have considered. What am I doing wrong?

Here was how I have approached this problem most recently:
30/50 = (1.02)^3x / (1.47)^x

Is it necessary to use change of base on the RHS top and bottom and then substitue back in? I ask because that is in the section which follows the one we are covering....

Here is a slightly different way to solve the equation: I take your last line:

30/50 = (1.02)^3x / (1.47)^x . Re-written in LaTex:

$\dfrac{30}{50} = \dfrac{(1.02)^{3x}}{(1.47)^x}$

$\dfrac{30}{50} = \left(\dfrac{(1.02)^{3}}{(1.47)}\right)^x$ And now you have to use logarithms:

$\log(0.6)=x\cdot \log\left(\dfrac{(1.02)^{3}}{(1.47)}\right)$

I'll leave the rest for you.
• Oct 7th 2009, 10:36 PM
strngFan
Thank you earboth!
Thank you for showing me that I was pursing a reasonable direction for solving the problem but failed in execution when I did not recognize that I could raise the RHS fraction to the power of x in order to apply the law of logarithms there.

Very nice solution!!

Sarah