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If $$x = 2\log_39 + \log_{27}5,$$ then $3^x = ??$

I know $2\log_39$ is $4$, however how do I find an exact answer to the whole equation??

I tried to change the base of the second one to $ 3 $, however it did not work out.

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$$\begin{align}x= 2\log_{3}9+\log_{27}5\\ =2\log_{3}9+{1\over3}\log_{3}5 \end{align}$$

$$\begin{align}3^x=3^{[2\log_{3}9+1/3\log_{3}5]}& &\\=3^{2\log_{3}9}3^{1/3\log_{3}5} &\\=9^{2\log_{3}3}5^{1/3\log_{3}3} \\=9^{2}\cdot 5^{1/3} \\=81\sqrt[3]5 \end{align}$$

we use :

$a^{\log(b)}=b^{\log(a)}$

proof:

Let :$ x = a^{\log(b)}$, and $y = b^{\log(a)}$

$\log x = \log{(a^{\log(b)})} = \log(b)\log(a) $

$\log y = \log{(b^{\log(a)})} = \log(a)\log(b) $

$\log(a)\log(b) = \log(b)\log(a)$

$ \log x = \log y$

$x = y $

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  • 1
    $\begingroup$ +1; use \log x$(\log x)$ instead of log$(log)$. $\endgroup$ – A---B Dec 27 '16 at 21:36
  • $\begingroup$ @A---B Thanks . $\endgroup$ – W.R.P.S Dec 27 '16 at 21:43
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    $\begingroup$ Check if my edit is ok or not. $\endgroup$ – A---B Dec 27 '16 at 21:47
4
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Hint:

$$\log_{27}5=\frac{\ln5}{\ln 27}=\frac{\ln 5}{3\ln 3}=\frac13\log_3 5$$

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You must write $\log_{27}(5) = \frac{\log_3(5)}{\log_3(27)} = \frac{\log_3(5)}{3} = \log_3(5^{1/3})$

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3
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we write $$x=2\log_3 3^2+\frac{\ln(5)}{3\ln(3)}=4+\frac{1}{3}\log_3 5$$

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3
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$$\log_{27} 5 = \frac{\log_3 5}{\log_{3} 27},$$ using the change of base formula $$\log_b x = \frac{\log_c x}{\log_c b}.$$ Now can you proceed?


Proof of the change of base formula:

Note that if there is a number $y$ such that $\log_b x = y$, then $b^y = x$. Now taking base-$c$ logarithms of both sides, we get $$\log_c b^y = \log_c x.$$ Now the LHS can be written $\log_c b^y = y \log_c b$, hence $$y \log_c b = \log_c x,$$ therefore $$y = \frac{\log_c x}{\log_c b},$$ as claimed.

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Use the fact that $\log_{27}(5) = \frac{\log_3(5)}{\log_3(27)} = \frac{\log_3(5)}{3} = \log_3(5^{\frac{1}{3}})$.

Then $x = 2\log_3(9) + \log_{27}(5) = \log_3(81) + \log_3(5^\frac{1}{3}) = \log_3(81 \cdot 5^{\frac{1}{3}})$.

You obtain that $3^x = 81 \cdot 5^\frac{1}{3}$.

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