Okay, so here is what I have $\displaystyle P=1045ln(0.0016f+1)$

P= perceived pitch (mels), F= frequency (hertz)

I need to solve when P= 1500 algebraically.

Can anyone give me some sparks to start my engine? Thanks.

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- Apr 20th 2008, 08:00 AMItsbaxagainSolving an Exponetial & Log Equation.
Okay, so here is what I have $\displaystyle P=1045ln(0.0016f+1)$

P= perceived pitch (mels), F= frequency (hertz)

I need to solve when P= 1500 algebraically.

Can anyone give me some sparks to start my engine? Thanks. - Apr 20th 2008, 08:04 AMMoo
Hello,

Well, divide each side of the equation by 1045, then take the exponential of the two sides, and ... - Apr 20th 2008, 08:18 AMItsbaxagain
Well after doing that and reducing I got: $\displaystyle \frac{300}{209}=ln(.0016f+1)$ I don't know how to break up the ln part. Do I need it in log for like $\displaystyle \frac{300}{209}=\frac{log(.0016f+1)}{log(e)}$ but still I don't know how to break apart the $\displaystyle log(.0016f+1)$

- Apr 20th 2008, 08:21 AMMoo
If ln is the neperian one :

$\displaystyle \ln(0.016f+1)=300/209 \Longleftrightarrow 0.016f+1=e^{300/209}$

If it's the logarithm in basis 10 :

$\displaystyle \ln(0.016f+1)=300/209 \Longleftrightarrow 0.016f+1=10^{300/209}$ - Apr 20th 2008, 08:32 AMItsbaxagain
I confused, I can't see how I get the answer because when I find the answer using my calculator I get approx:1278.84

- Apr 20th 2008, 08:34 AMMoo
I should have added "then, isolate f " (Thinking)

- Apr 20th 2008, 08:49 AMItsbaxagain