2
$\begingroup$

The equation is

$64(12)^{8x} = 195$

the steps I have done so far.

  • $12^{8x} = 195/64$
  • $8x \cdot \ln(12) = \ln(195/64)$
  • $8x = (\ln(195/64))/(\ln(12))$

not sure how to divide $8x$ to get $x$ by itself for the final answer with the ln's

  • 6
    Wait a minute --- you've done all the log stuff, now you're down to $8x=A$, and you can't solve that for $x$?2012-12-04
  • 0
    @GerryMyerson that is correct. I took a break from math for about 8 months and I know you divide 8 on both sides, but the ln's are confusing me.2012-12-04
  • 0
    Don't let them confuse you! Dividing by $8$ is dividing by $8$, no matter what the other side of the equation looks like. What would you do if it were $8x=93/42$?2012-12-04
  • 0
    @GerryMyerson x = 8(93/42)?2012-12-04
  • 0
    @GerryMyerson oh its x = 93/3362012-12-04
  • 0
    I've corrected a typo on the RHS of your 2nd step. It was $(195/64)\ln$.2012-12-04
  • 0
    Yes, and that $336$ was just $8\times42$, right? And you could have written $(8)(42)$ instead of $336$, right? So, ${93\over(8)(42)}$, right? So, can you finish the log problem off now?2012-12-04
  • 0
    @GerryMyerson Yup! thank you2012-12-05

1 Answers 1

3

when you divide $8x = (\ln(195/64))/\ln(12)$ you get $x = (\ln(195/64))/(\ln(12)(8))$. The 8 only affects the bottom denominator of equation when you divide a fraction. thank you @GerryMyerson

  • 0
    It's better to write $x=\frac{\ln(195/64)}{8\ln(12)}$ using the code `$x=\frac{\ln(195/64)}{8\ln(12)}$`. See [MathJax basic tutorial and quick reference](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference).2012-12-05
  • 0
    @AméricoTavares thanks, I don't understand how that code works yet. Do you know of any great tutorials?2012-12-05
  • 0
    For some basic information about writing math at this site see e.g. [here](http://meta.math.stackexchange.com/questions/5020/), [here](http://meta.stackexchange.com/a/70559/155238), [here](http://meta.math.stackexchange.com/questions/1773/) and [here](http://math.stackexchange.com/editing-help#latex).2012-12-05