Logarithmic expressions

**dan713** · August 6th 2012, 10:01 PM

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

**richard1234** · August 6th 2012, 10:13 PM

Note that $\log_8 {64} = 2$ . Also, $\log_8 {64} = 3 \log_8 {4}$ by a logarithmic identity. Therefore

$3 \log_8 {4} = 2 \Rightarrow \log_8 {4} = \frac{2}{3}$ .

**dan713** · August 7th 2012, 08:03 AM

There is no exponent 3 though. How did you end up with log64 base 8?

**richard1234** · August 7th 2012, 08:16 AM

I'm using 64 because $64 = 4^3 = 8^2$ .

**Plato** · August 7th 2012, 08:43 AM

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.
$x = {\log _8}(4)\; \Rightarrow {8^x} = 4\; \Rightarrow \;x = \frac{2}{3}$ .

**richard1234** · August 7th 2012, 08:51 AM

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 $2^{3x} = 2^2$ .

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