Rewriting Logs

**x5pyd3rx** · February 20th 2010, 10:22 AM

I am a little confused on how to rewrite this

3lnx - 4lnx - 5lnt

it dosent seem right to right it as

$\ln \frac{\frac{ x^3}{x^4}}{ t^5}$

**Keithfert488** · February 20th 2010, 10:34 AM

That actually is correct. Just simplify. $\frac{x^3}{x^4}=\frac{1}{x}$

So its just $\ln \frac{1}{xt^5}$

Or you could just combine like terms before attempting to rewrite it.
$3\ln x-4\ln x=-\ln x$

**e^(i*pi)** · February 20th 2010, 10:35 AM

Originally Posted by x5pyd3rx

I am a little confused on how to rewrite this

3lnx - 4lnx - 5lnt

it dosent seem right to right it as

$\ln \frac{\frac{ x^3}{x^4}}{ t^5}$

That's right but it's poor syntax

$\ln(x^3) - \ln(x^4) - \ln(t^5) = \ln \left(\frac{x^3}{x^4t^5}\right) = \ln \left(\frac{x^{-1}}{t^5}\right) = -\ln \left({xt^{5}}\right)$

**Raoh** · February 20th 2010, 10:36 AM

Originally Posted by x5pyd3rx

I am a little confused on how to rewrite this

3lnx - 4lnx - 5lnt

it dosent seem right to right it as

$\ln \frac{\frac{ x^3}{x^4}}{ t^5}$

hi

$3\ln x - 4\ln x - 5\ln t=(\ln x^3-\ln x^4)-\ln t^5=\ln(\frac{x^3}{x^4})-\ln t^5$

**skeeter** · February 20th 2010, 10:37 AM

Originally Posted by x5pyd3rx

I am a little confused on how to rewrite this

3lnx - 4lnx - 5lnt

it dosent seem right to right it as

$\ln \frac{\frac{ x^3}{x^4}}{ t^5}$

$3\ln{x} - (4\ln{x} + 5\ln{t})$

$\ln(x^3) - \ln(x^4 t^5)$

$\ln\left(\frac{x^3}{x^4 t^5}\right)$

$\ln\left(\frac{1}{x t^5}\right)$

$-\ln(x t^5)$

**x5pyd3rx** · February 20th 2010, 10:40 AM

Crap i just got all this help and now i feel stupid because i wrote the problem wrong

its 3lnx - 4lny -5lnt

**e^(i*pi)** · February 20th 2010, 10:41 AM

Originally Posted by x5pyd3rx

Crap i just got all this help and now i feel stupid because i wrote the problem wrong

its 3lnx - 4lny -5lnt

It's pretty much equal to skeeter's second step

$\ln \left(\frac{x^3}{y^4 t^5}\right)$

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