I'm solving logs question. And got stuck in this question.

Given that $\log_2 3 = a$, $\log_3 5 = b$ and $\log_7 2 = c$, express the logarithm of number 63 to the base 140 in terms of a,b and c

I've tried to sole it. But wasn't able to move after it. enter image description here

Please explain, how I can solve it and please solve it according to class 11th student.

  • $\begingroup$ Do you know base change formulas for logarithms? $\endgroup$ – rhombicosicodecahedron May 1 '19 at 20:19
  • $\begingroup$ Yes, I've tried to break base in 140. But wasnt able to break $\log_{140} 2$ in denominator $\endgroup$ – Piyush Raj May 1 '19 at 20:23
  • $\begingroup$ $\log_a(b) = \frac{\log_c(b)}{\log_c(a)}$ in any base $c$ and $\log(140) = \log(2^2\cdot 5\cdot 7) = 2\log(2) + \log(5) + \log7$ $\endgroup$ – achille hui May 1 '19 at 20:24

A way to make our lives somewhat more difficult: $$\log_23=a\Rightarrow2^a=3\\\log_35=b\Rightarrow3^b=5\\\log_72=c\Rightarrow7^c=2\Rightarrow 7=2^{\frac1c}$$

Let $x=\log_{140}63\Rightarrow140^x=63$

Notice that $140=20*7=2*2*5*7$ and $63=7*9=3*3*7$

So we write everything dependent on $2, 5$ and $c$. $$(2*2*5*7)^x=2^a*2^a*2^{\frac1c}\Rightarrow\\(2*2*5*2^{\frac1c})^x=2^a*2^a*2^{\frac1c}\Rightarrow\\x\log_2(2^{\frac1c+2})+\log_25=2a+\frac1c\Rightarrow\\x=c\frac{[2a-\log_25]+1}{2c+1}$$

Technically we get rid of $b$, though we have the $\log_25$ in the expression..


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