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

Question

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! :)

Answer 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..

Answer 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?

Answer 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.

Answer 4

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