I am a beginner to logarithms. By using logarithms, in this equation, I want to find the value of $x$:
$$ \ 3^x = 4^{x-1} \ $$
I am using 10 as base of logarithm. And I want $x$ in terms of $\log2$ and $\log3$, as far as possible.
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4If you are using $10$ as your base of logarithm, why do you want to use $\ln 2$ and $\ln 3$, which are the natural logarithms (base $e$?) – Thomas Andrews Jun 29 '17 at 17:44
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Note that $(4/3)^x=4$. – MAN-MADE Jun 29 '17 at 17:58
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@Thomas Andrews It was a silly typing error.. sorry.. I've now edited it – Fghj Jun 29 '17 at 18:09
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Hint: Whichever base $b$ you use for the logarithm, we have $$ \log_b(a^k) = k \log_b(a) $$ for any $a,k$. Take a logarithm of both sides of your equation.
Ben Grossmann
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As mentioned logarithm use is desired. Here, for a solution in terms of natural logarithms, \begin{align} \ln(3^{x}) &= \ln(4^{x-1}) \\ x \, \ln(3) &= (x-1) \, \ln(4) \\ x \, (\ln(3) - \ln(4)) &= - \ln(2^{2}) = - 2 \, \ln(2) \\ x &= - \frac{2 \, \ln(2)}{\ln\left(\frac{3}{4}\right)} \end{align}
Converting from natural logarithm to a base 10 logarithm:
Use $$\log_{b}(x) = \frac{\log_{d}(x)}{\log_{d}(b)}$$ to convert the base of the logarithms. Here, let $b = e$ and $d = 10$ to yield \begin{align} x = - 2 \, \cdot \frac{\log_{10}(2)}{\log_{10}(e)} \, \frac{\log_{10}(e)}{\log_{10}\left(\frac{3}{4}\right)} = -2 \, \frac{\log_{10}(2)}{\log_{10}\left(\frac{3}{4}\right)}. \end{align}
Strictly solving by base 10 logarithm: \begin{align} \log_{10}(3^{y}) &= \log_{10}(4^{y-1}) \\ y \, \log_{10}(3) &= y \, \log_{10}(4) - \log_{10}(4) \\ y \, (\log_{10}(3) - \log_{10}(4)) &= - \log_{10}(4) \\ y &= \frac{- \, \log_{10}(4)}{\log_{10}(3) - \log_{10}(4)} = \frac{2 \, \log_{10}(2)}{\log_{10}\left(\frac{4}{3}\right)}. \end{align}
Leucippus
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4@RaviPrakash: then this should be mentioned when you ask the problem. – Bob Krueger Jun 29 '17 at 17:41
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you can simplify the expression above to get what you want pretty easily; remember that $$ln(4) = ln(2^2) = 2ln(2)$$ – Rohil Verma Jun 29 '17 at 17:46
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Don't re-use the value $x$ in your formula starting the base $10$ case, it is apt to be confusing. – Thomas Andrews Jun 29 '17 at 18:13