I am a little confused on how to rewrite this 3lnx - 4lnx - 5lnt it dosent seem right to right it as $\displaystyle \ln \frac{\frac{ x^3}{x^4}}{ t^5}$
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That actually is correct. Just simplify. $\displaystyle \frac{x^3}{x^4}=\frac{1}{x}$ So its just $\displaystyle \ln \frac{1}{xt^5}$ Or you could just combine like terms before attempting to rewrite it. $\displaystyle 3\ln x-4\ln x=-\ln x$
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 $\displaystyle \ln \frac{\frac{ x^3}{x^4}}{ t^5}$ That's right but it's poor syntax $\displaystyle \ln(x^3) - \ln(x^4) - \ln(t^5) = \ln \left(\frac{x^3}{x^4t^5}\right) = \ln \left(\frac{x^{-1}}{t^5}\right) = -\ln \left({xt^{5}}\right)$
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 $\displaystyle \ln \frac{\frac{ x^3}{x^4}}{ t^5}$ hi $\displaystyle 3\ln x - 4\ln x - 5\ln t=(\ln x^3-\ln x^4)-\ln t^5=\ln(\frac{x^3}{x^4})-\ln t^5$
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 $\displaystyle \ln \frac{\frac{ x^3}{x^4}}{ t^5}$ $\displaystyle 3\ln{x} - (4\ln{x} + 5\ln{t})$ $\displaystyle \ln(x^3) - \ln(x^4 t^5)$ $\displaystyle \ln\left(\frac{x^3}{x^4 t^5}\right)$ $\displaystyle \ln\left(\frac{1}{x t^5}\right)$ $\displaystyle -\ln(x t^5)$
Crap i just got all this help and now i feel stupid because i wrote the problem wrong its 3lnx - 4lny -5lnt
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 $\displaystyle \ln \left(\frac{x^3}{y^4 t^5}\right)$
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