1
$\begingroup$

Given :

$$\log_{12}18 = a \text{ and }\log_{24}54=b$$

prove that:

$$ab + 5(a-b) = 1$$

My attempt: I couldn't solve it in any way, as base were not common. I could solve it if base of second equation was $12$ raised to something, but here it is $12\times 2$ :( Any help will be greatly appreciated.

$\endgroup$
1
  • $\begingroup$ ..."prove this logarithm"?? Huh? $\endgroup$ May 13, 2015 at 16:49

2 Answers 2

3
$\begingroup$

HINT:

$$a=\log_{12}18=\dfrac{\log(18)}{\log(12)}=\dfrac{\log2+2\log3}{2\log2+\log3}$$

as $\log_ab=\dfrac{\log b}{\log a}$ and $\log(a^2b)=\log(a^2)+\log(b)=2\log a+\log b$

Express $\log2$ in terms of $\log3$

Similarly for $b$

Then compare the values of $\log2$ to eliminate it

$\endgroup$
3
3
$\begingroup$

Note that $a=\frac{log(18)}{log(12)}$ and $b=\frac{log(54)}{log(24)}$

$\endgroup$
1
  • $\begingroup$ Oh it was so simple! many thanks! $\endgroup$
    – Max Payne
    May 13, 2015 at 16:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .