2log(2x) = 1 + log(a)

orangeiv · #1 $Old$ October 16th, 2009, 11:41 AM

I'm having a problem manipulating some logs...I've only just started using them and finding them a pain, although I'm assured they're beautiful things once you get the hang of them.

Anyway here's a question I'm having the problem with:

Solve x in terms of a, where a>0

2log(2x) = 1 + log(a)

I know the answer is (5a/2)^0.5 but I just can't work out how to get there

Defunkt · #2 $Old$ October 16th, 2009, 12:29 PM

Quote:

Originally Posted by orangeiv $View Post$

I'm having a problem manipulating some logs...I've only just started using them and finding them a pain, although I'm assured they're beautiful things once you get the hang of them.

Anyway here's a question I'm having the problem with:

Solve x in terms of a, where a>0

2log(2x) = 1 + log(a)

I know the answer is (5a/2)^0.5 but I just can't work out how to get there

$2log(2x) = log((2x)^2)$

$1 + log(a) = log(e) + log(a) = log(ea)$

===> $log(4x^2) = log(ea)$

Can you finish now?

earboth · #3 $Old$ October 16th, 2009, 12:30 PM

Quote:

Originally Posted by orangeiv $View Post$

I'm having a problem manipulating some logs...I've only just started using them and finding them a pain, although I'm assured they're beautiful things once you get the hang of them.

Anyway here's a question I'm having the problem with:

Solve x in terms of a, where a>0

2log(2x) = 1 + log(a)

I know the answer is (5a/2)^0.5 but I just can't work out how to get there

1. Separarte logs and constants on different sides of the equation:

$2\log(2x)-\log(a)=1$

2. Use all the log-rules you know to simplify the LHS:

$\log\left(\dfrac{4x^2}{a}\right)=1$

3. Now transform into a exponential equation:

$10^{\log\left(\frac{4x^2}{a}\right)}= 10^1$

4. Solve for x:

$\dfrac{4x^2}{a} = 10$

5. I'll leave the rest for you.