$Let$ $a,b,c>0$



So my attempt is just to use the logarithmic definition:

$$\log_a(b^x)=2\iff b^x=a^2$$

By similar logic,

$$a^x=c^5$$ $$c^x=b^2$$

So, if add everything together, we should get:


Looks to me like x is equal 2 different numbers at the same time which is strange, what am I doing wrong here?

I'm going to be a maths student in the upcoming year, this is taken from the Tel-Aviv university preparation material - shouldn't be too complex.

  • $\begingroup$ What do you mean by "Let $a,b,c>0$"? Are they known/constant, or unknown/variable? If they are known, then you have an over-determined system of 3 equations with 1 unknown. For example, if $a=b=c=7$, then the first equation gives $x=2$, but the third gives $x=5\neq2$. $\endgroup$ – mr_e_man Jun 22 '18 at 4:19
  • $\begingroup$ I think you have a typo: $\log_b(c^x)=2$ but $c^x=b^3$. Is it $2$ or $3$? $\endgroup$ – Burnsba Jun 22 '18 at 13:13
  • $\begingroup$ I have just checked and it was actually a 3 but for the sake of all the nice answers I've recieved, I'll keep it a 2, it doesn't really matter anyway. $\endgroup$ – L0wRider Jun 22 '18 at 19:45

Needless to use the natural logarithm: from $$\begin{cases} \log_a b^x=2 \\\log_b c^x=2 \\\log_ca^x=5 \end{cases} \iff \begin{cases} x\log_a b=2 \\x\log_b c=2 \\x\log_ca=5 \end{cases} $$ you deduce readily that $$x^3(\underbrace{\log_ab\,\log_b c\,\log_c a}_{=1})=20.$$

  • 3
    $\begingroup$ \underbrace… fancy! 😀 $\endgroup$ – André Levy Jun 22 '18 at 0:22
  • $\begingroup$ You could just use one equation, to get $x = 2\log_b a$, or $x = 5\log_a c$, if the parameters $a,b,c$ are known. $\endgroup$ – mr_e_man Jun 22 '18 at 4:17
  • $\begingroup$ @mr_e_man: That's right, but I thought more elegant to do the calculation generically, with just one well-known relation between logarithms of different bases. $\endgroup$ – Bernard Jun 22 '18 at 7:33
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    $\begingroup$ Although the relation is simple, I have not recognized it at the first glance, so I'd like to see it expanded through a basic identity $\log_ab=\frac{\log b}{\log a}$ which leads to a 6-term fraction reducing to 1/1. $\endgroup$ – CiaPan Jun 22 '18 at 9:29
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    $\begingroup$ This formula is standard as an exercise on logarithms with different bases — there aren't so many exercises on this theme. $\endgroup$ – Bernard Jun 22 '18 at 9:58

Using change of base and power formulea, we have \begin{eqnarray*} x \ln b = 2 \ln a \\ x \ln c= 2 \ln b \\ x \ln a = 5 \ln c \\ \end{eqnarray*} So \begin{eqnarray*} x^3 \ln a = 5 x^2 \ln c= 10 x \ln b =20 \ln a \\ \end{eqnarray*} Should be easy from here ?


$\displaystyle\begin{cases}\log_a(b^x)=2\\log_b(c^x)=2\\log_c(a^x)=5\end{cases}\implies \begin{cases}\frac{\ln(b^x)}{\ln(a)}=2\\\frac{\ln(c^x)}{\ln(b)}=2\\\frac{\ln(a^x)}{\ln(c)}=5\end{cases}$


adding them all together gives :


$\implies x\bigg(\ln(b)+\ln(a)+\ln(c)\bigg)=2\ln(a)+5\ln(c)+2\ln(b) $

$\implies x= \dfrac{2\ln(a)+5\ln(c)+2\ln(b)}{\ln(b)+\ln(a)+ln(c)}$


The power of logarithm can be transferred to the front as a coefficient

$$ \log_a b^x=2 \rightarrow \,x\log_a b= \frac{x\cdot\log b}{\log a}=2 $$ irrespective of any chosen base, need even not be mentioned ( so long as it is real);

$$ \frac{x\cdot\log b}{\log a}\cdot\frac{x\cdot\log c}{\log b}\cdot\frac{x\cdot\log a}{\log c}= 2\cdot 2\cdot 5= 20$$

$$ x^3=20,\, x= 20^{1/3}. $$


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