1
$\begingroup$

Given that $\log12=1.0792$ and $\log4=0.6021$, solve $\log8$ without a calculator.

I am familiar with the following three rules:

  1. Product rule: $\log(a\cdot b)=\log a+\log b$
  2. Quotient rule: $\log(a/b)=\log a-\log b$
  3. Power rule: $\log(a^b)=b\cdot\log a$

But I honestly don't see how they help in this case. Any way you slice it, it seems like it's necessary to introduce $\log2$, which is not given. Am I missing something?

$\endgroup$
  • 1
    $\begingroup$ You can introduce $\log(2)$ easily. $\log(4) = 2\log(2)= 0.6021 $. $\endgroup$ – Parth Kohli Jun 5 '14 at 5:36
3
$\begingroup$

$\log 4 = \log 2^2 =2\log 2 =0.6021$
$\implies\log 2 =0.3010$
$\log 8 = \log 2^3 = 3\log 2 = 3*0.3010 = 0.9030$

$\endgroup$
3
$\begingroup$

Hint: $8^2=4^3$.

Alternative hint: $2=\sqrt4\,$.

$\endgroup$
2
$\begingroup$

$\log 8=\log(4^{3/2})$

Use the $3^{\text{rd}}$ rule.

$\endgroup$
1
$\begingroup$

Since you received the answers, let me be slightly more general and show you what you can do just knowing that $\log_{10}(12)=1.0792$ and $\log_{10}(4)=0.6021$. Since $4=2^2$, you then have $\log_{10}(2)=0.3010$. Since $3=\frac{12} {4}$, you then have $\log_{10}(3)=0.4771$. We could continue like that for quite many numbers.

But, by the end, wht this means is that, just based on this mimited information, you can compute the logarithm of any number which write $2^a3^b4^c6^d8^e9^f10^g$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.