Log fraction with no calculator

**desiderius1** · April 13th 2010, 11:12 PM

How would I solve $\log_3 {\frac94} + \log_3 4$ without using a calculator.

This is far as I got:

$\frac{\log 9/4}{\log 3} + \frac{\log 4}{\log 3}$

Thanks!

**earboth** · April 13th 2010, 11:15 PM

Originally Posted by desiderius1

How would I solve $\log_3 {\frac94} + \log_3 4$ without using a calculator.

This is far as I got:

$\frac{\log 9/4}{\log 3} + \frac{\log 4}{\log 3}$

Thanks!

$\log_3 \left({\frac94}\right) + \log_3 (4) = \log_3(9)-\log_3(4)+\log_3(4)$

Simplify! Keep in mind that $\log_3(9)=2$

**desiderius1** · April 14th 2010, 01:08 AM

Thank you.

I have another log question. Cannot use calculator.

$\log_6 8 + \log_6 9 - \log_6 2$

So, should I add or subtract first?

if I add $\log_6 (8*9)$ I get $\log_6 (72)$ which I can't figure out how to simplify any further. If I subtract $\log_6 \frac{9}{2}$ , what would I do next?

Thanks for helping.

**desiderius1** · April 14th 2010, 01:53 AM

Ah, ok I figured it out.

$log_6 8 + log_6 9 = log_6 (8*9) = log_6 (72)$
$log_6 \frac{72}{2} = log_6 36<br /> = log_6 6^2 = 2$

**desiderius1** · April 14th 2010, 02:30 AM

more log questions:

$log_6 4 - log_6 3 - log_6 8$

so far I've done:

$log_6 4 - log_6 8 = log_6 {\frac48} = log_6 {\frac12}$
$log_6{\frac12} - log_6 3$
Stuck here.

Other problems...
$log_7 \frac{3}{147}$

$log_2(log_4 2 + log_4 8)^3$

$3(4log_4 12-2log_4 9)$

thanks!~

**earboth** · April 14th 2010, 04:39 AM

Originally Posted by desiderius1

more log questions:

$log_6 4 - log_6 3 - log_6 8$

so far I've done:

$log_6 4 - log_6 8 = log_6 {\frac48} = log_6 {\frac12}$
$log_6{\frac12} - log_6 3$
Stuck here.

Re-write the difference of logs as a log of a quotient:

$\log_6\left(\frac12\right) - \log_6( 3)= \log_6\left(\dfrac{\frac12}{3}\right)=\log_6 \left(\frac16 \right) = \log_6(6^{-1})$

Continue!

Other problems...
$log_7 \frac{3}{147}$

$log_2(log_4 2 + log_4 8)^3$

$3(4log_4 12-2log_4 9)$

thanks!~

To help you sufficiently I must know what you have done so far and where you have difficulties doing these questions.
Please show your work.

**Dimple** · April 14th 2010, 06:35 AM

Originally Posted by desiderius1

Other problems...
$log_7 \frac{3}{147}$

Try log7 ((147/3)^-1).

Edit:

Originally Posted by desiderius1

$<br /> log_2(log_4 2 + log_4 8)^3<br />$

log4 2 <=> 4^x = 2 and log4 8 <=> 4^y = 8. Solve x and y first. Makes it a lot easier.

**e^(i*pi)** · April 14th 2010, 06:42 AM

Originally Posted by desiderius1

more log questions:

$log_6 4 - log_6 3 - log_6 8$

so far I've done:

$log_6 4 - log_6 8 = log_6 {\frac48} = log_6 {\frac12}$
$log_6{\frac12} - log_6 3$
Stuck here.

Other problems...
$log_7 \frac{3}{147}$

$log_2(log_4 2 + log_4 8)^3$

$3(4log_4 12-2log_4 9)$

thanks!~

Note that 3 will cancel: $\frac{3}{147} = \frac{1}{49}$ which is much easier to manipulate

**bjhopper** · April 14th 2010, 11:08 AM

[quote=desiderius1;493586]more log questions:

$log_6 4 - log_6 3 - log_6 8$

so far I've done:

$log_6 4 - log_6 8 = log_6 {\frac48} = log_6 {\frac12}$
$log_6{\frac12} - log_6 3$
Stuck here.

Other problems...
$log_7 \frac{3}{147}$

$log_2(log_4 2 + log_4 8)^3$

Log B2 {logB4 (2) +logB4(8) }^3

Work inside brackets first
!/2 +logB4(16/2)=1/2 +2 -1/2 = 2 2^3=8
Now you have logB2(8)=3

bjh

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