# quick log question...

• May 20th 2009, 09:56 AM
chinkmeista
quick log question...
Simplify \$\displaystyle 2loga+3logb\$:

I thought it you could do add the 2 and 3, but multiply a + b to give \$\displaystyle 5logab\$.

But the answer says: \$\displaystyle 2loga+3logb = log{a^2}+log{b^2} = log{a^2}{b^3}\$.

I perfectly understand how the mark scheme did it, moving the powers of 2 and 3 form the log laws, but i was just wondering though:

Is it wrong adding the first bit I did of the 2 and the 3 before the logs, and multiplying the a and b?

Thanks.
• May 20th 2009, 10:11 AM
e^(i*pi)
Quote:

Originally Posted by chinkmeista
Simplify \$\displaystyle 2loga+3logb\$:

I thought it you could do add the 2 and 3, but multiply a + b to give \$\displaystyle 5logab\$.

But the answer says: \$\displaystyle 2loga+3logb = log{a^2}+log{b^2} = log{a^2}{b^3}\$.

I perfectly understand how the mark scheme did it, moving the powers of 2 and 3 form the log laws, but i was just wondering though:

Is it wrong adding the first bit I did of the 2 and the 3 before the logs, and multiplying the a and b?

Thanks.

Use the law \$\displaystyle klog(a) = log(a^k)\$

then you can use the law \$\displaystyle log(a) + log(b) = log(ab)\$
• May 20th 2009, 10:20 AM
chinkmeista
Quote:

Originally Posted by e^(i*pi)
Use the law \$\displaystyle klog(a) = log(a^k)\$

then you can use the law \$\displaystyle log(a) + log(b) = log(ab)\$

Oh ok, you can only do \$\displaystyle log(a) + log(b) = log(ab) \$ after that power rule?

If the question was 2log(a)+3log(a),can i add it then to 5log(a^2) because they're both 'a' or do i still have to use that power rule first and make it log(a^2) + log(a^3) = log(a^5)?
• May 20th 2009, 10:31 AM
e^(i*pi)
Quote:

Originally Posted by chinkmeista
Oh ok, you can only do \$\displaystyle log(a) + log(b) = log(ab)\$ after that power rule?

If the question was 2log(a)+3log(a),can i add it then to 5log(a^2) because they're both 'a' or do i still have to use that power rule first and make it log(a^2) + log(a^3) = log(a^5)?

You have to use the power law in your first example because your logs are not the same because 2 does not equal 3.

In your second example you can add them because a=a. Think about what happens if you use the power rule on log(a^5) (Wink)
• May 20th 2009, 10:40 AM
chinkmeista
Quote:

Originally Posted by e^(i*pi)
You have to use the power law in your first example because your logs are not the same because 2 does not equal 3.

In your second example you can add them because a=a. Think about what happens if you use the power rule on log(a^5) (Wink)

nice one, cheers buddy lol, ive actually spent a hour on this minor problem! lol thanks.