math

how do i solve for x when x = log(32.5)?

here's my work:

log x = y
10^y = x
then,
x = log (32.5)
10^x = 32.5
10^x = 17/2

this is where i get stuck... i don't know how to get rid of the bases... can some one please teach me how? thank you! :)

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  1. ooopsss i think i know where i went wrong. it's supposed to be 10^x = 65/2 and then i apply the natural log to both sides.... lemme check..

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  2. so here's my edited work:

    10^x = 65/2
    xln10 = ln (65/2)
    xln10 = ln65 - ln2
    x = ln65 - ln2 - ln 10
    x = 1.17865~ , BUT

    when i check the original equation: x = log 32.5, x = 1.51188~

    hmmm they don't match up. can someone please tell me where i'm going wrong?

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  3. what's to solve? x = log 32.5 = 1.51188

    The base is 10.

    log 32.5 is the power of 10 you need to get 32.5. So,

    10^(log 32.5) = 32.5
    if x = log 32.5, then 10^x = 32.5

    That's it. There is no further to go. Log and exponent are inverse operations, just like + and -, * and /, sqrt and ^2.

    If x = sqrt(10), then x^2 = 10

    As for your math above, if you want to change bases,

    xln10 = ln65-ln2
    ln10 is just a number, like 2.3, so
    x = (ln65-ln2)/ln10
    no subtraction involved

    If you had had
    10x = ln65 - ln2, then that would have been
    ln10 + lnx = ln65 - ln2
    and your subtraction would have been correct.

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  4. actually, a typo:
    ln(10x) = ln65 - ln2

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