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! :)
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..
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?
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.
actually, a typo:
ln(10x) = ln65 - ln2
To solve for x when x = log(32.5), you can use the property of logarithms that states log_b(a) = c is equivalent to b^c = a. In your case, you have x = log(32.5), which can be rewritten as 10^x = 32.5.
To get rid of the base 10, you can take the logarithm of both sides using the base 10 logarithm (common logarithm). This means you can apply the property log_b(b^c) = c, which in this case would be log(10^x) = log(32.5).
Using the property log_b(b^c) = c, the left side simplifies to x, so you have x = log(32.5).
To evaluate log(32.5), you can use a calculator or a logarithmic table. The value of log(32.5) is approximately 1.5114 (rounded to four decimal places). Therefore, x ≈ 1.5114.
So, x is approximately equal to 1.5114 when x = log(32.5).