# Logarithmic expressions

• August 6th 2012, 10:01 PM
dan713
Logarithmic expressions
Hi, I need help with the following question in the attachment. It is about logs, I don't understand this problem. Please help, thanks in advanced :)
• August 6th 2012, 10:13 PM
richard1234
Re: Logarithmic expressions
Note that $\log_8 {64} = 2$. Also, $\log_8 {64} = 3 \log_8 {4}$ by a logarithmic identity. Therefore

$3 \log_8 {4} = 2 \Rightarrow \log_8 {4} = \frac{2}{3}$.
• August 7th 2012, 08:03 AM
dan713
Re: Logarithmic expressions
There is no exponent 3 though. How did you end up with log64 base 8?
• August 7th 2012, 08:16 AM
richard1234
Re: Logarithmic expressions
I'm using 64 because $64 = 4^3 = 8^2$.
• August 7th 2012, 08:43 AM
Plato
Re: Logarithmic expressions
Quote:

Originally Posted by dan713
Hi, I need help with the following question in the attachment. It is about logs, I don't understand this problem. Please help, thanks in advanced :)

Here is another way to look at it.
$x = {\log _8}(4)\; \Rightarrow {8^x} = 4\; \Rightarrow \;x = \frac{2}{3}$.
• August 7th 2012, 08:51 AM
richard1234
Re: Logarithmic expressions
Quote:

Originally Posted by Plato
Here is another way to look at it.
$x = {\log _8}(4)\; \Rightarrow {8^x} = 4\; \Rightarrow \;x = \frac{2}{3}$.

That is correct but it might not be immediately apparent to dan713 that x = 2/3. Unless you note that $\sqrt[3]{8^2} = 4$, or you write it as $2^{3x} = 2^2$.