Logarithm Using Givens (without calculator)

• Jun 16th 2013, 02:52 PM
ReneG
Logarithm Using Givens (without calculator)
Given $\log 2 = 0.3010$ and $\log 3 = 0.4771$ find $\log_5{512}$

I was only able to simplify to $9\log_5{2}$, but I don't know what to do. I tried using the change of base formula, but that left me with $\log 5$ which I couldn't compute algebraically
• Jun 16th 2013, 02:59 PM
Plato
Re: Logarithm Using Givens (without calculator)
Quote:

Originally Posted by ReneG
Given $\log 2 = 0.3010$ and $\log 3 = 0.4771$ find $\log_5{512}$

I was only able to simplify to $9\log_5{2}$, but I don't know what to do. I tried using the change of base formula, but that left me with $\log 5$ which I couldn't compute algebraically

I think that there is a typo in the statement of the question.

I think the 5 should be a 3.
• Jun 16th 2013, 03:01 PM
ReneG
Re: Logarithm Using Givens (without calculator)
I wish that was the case, but it's not.
• Jun 16th 2013, 03:11 PM
Plato
Re: Logarithm Using Givens (without calculator)
Quote:

Originally Posted by ReneG
I wish that was the case, but it's not.

Then that problem cannot be done. But I am willing to bet it is a typo.
• Jun 16th 2013, 07:13 PM
ReneG
Re: Logarithm Using Givens (without calculator)
I've figured it out $\log_5{512} = \frac{9\log 2}{\log{10} - \log 2} = \frac{9(0.3010)}{1-0.3010}$

I didn't need log 3 to solve it though.
• Jun 16th 2013, 07:33 PM
topsquark
Re: Logarithm Using Givens (without calculator)
Quote:

Originally Posted by ReneG
I didn't need log 3 to solve it though.

Nice work! However since you didn't need the log(3) I'm skeptical...

-Dan