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claire3158 · Jul 27, 2009, 05:25 PM

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I have 4 logarithmic problems that I need help with...

log10 (10^1/2)

log10 (1/10^x)

log 3 x^2=2log3 4- 4log3 5

2 log4 9- log2 3

galactus · Jul 28, 2009, 03:58 PM

$2 log_{4}(9)- log_{2}(3)$

Try using the change of base formula. $log_{a}b=\frac{log(b)}{log(a)}$

$2\frac{log(9)}{log(4)}-\frac{log(3)}{log(2)}$

Property of logs:

$\frac{4log(3)}{2log(2)}-\frac{log(3)}{log(2)}$

Now, see it?

Unknown008 · Jul 29, 2009, 10:53 AM

$log_{10} 10^{\frac12} = \frac12 log_{10}10$

From property of logs. Can you simplify now?

$log_{10} \frac{1}{10^x}$

Use the property of logs; $log\frac ab = loga - logb$

to give:

$log_{10} \frac{1}{10^x} = log_{10}1 - log_{10}10^x$

Then, simplify to $log_{10}1 - log_{10}10^x = log_{10}1 - xlog_{10}10$

Can you see it now?

$log_3 x^2=2log_3 4- 4log_3 5$

Use law of subtraction of logarithms, which I told you earlier. You'll have firstly:

$log_3 x^2=log_3 4^2- log_3 5^4$

then;

$log_3 x^2=log_3 \frac{4^2}{5^4}$

The logarithms cancel each other, leaving behind:

$x^2=\frac{4^2}{5^4}$

Can you solve it now?