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Math Help - [SOLVED] Reviewing Logs and stuck on a problem...help?!?

  1. #1
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    [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....

    Please help if you can, thank you.
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  2. #2
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    Hello strngFan

    Welcome to Math Help Forum!
    Quote Originally Posted by strngFan View Post
    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....

    Please help if you can, thank you.
    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?

    Grandad
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  3. #3
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    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
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  4. #4
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    Quote Originally Posted by strngFan View Post
    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....

    Please help if you can, thank you.
    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.
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  5. #5
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    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
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