How would I solve $\displaystyle \log_3 {\frac94} + \log_3 4$ without using a calculator.
This is far as I got:
$\displaystyle \frac{\log 9/4}{\log 3} + \frac{\log 4}{\log 3}$
Thanks!
Thank you.
I have another log question. Cannot use calculator.
$\displaystyle \log_6 8 + \log_6 9 - \log_6 2$
So, should I add or subtract first?
if I add $\displaystyle \log_6 (8*9)$ I get $\displaystyle \log_6 (72)$ which I can't figure out how to simplify any further. If I subtract $\displaystyle \log_6 \frac{9}{2}$, what would I do next?
Thanks for helping.
more log questions:
$\displaystyle log_6 4 - log_6 3 - log_6 8$
so far I've done:
$\displaystyle log_6 4 - log_6 8 = log_6 {\frac48} = log_6 {\frac12}$
$\displaystyle log_6{\frac12} - log_6 3$
Stuck here.
Other problems...
$\displaystyle log_7 \frac{3}{147}$
$\displaystyle log_2(log_4 2 + log_4 8)^3$
$\displaystyle 3(4log_4 12-2log_4 9)$
thanks!~
Re-write the difference of logs as a log of a quotient:
$\displaystyle \log_6\left(\frac12\right) - \log_6( 3)= \log_6\left(\dfrac{\frac12}{3}\right)=\log_6 \left(\frac16 \right) = \log_6(6^{-1})$
Continue!
To help you sufficiently I must know what you have done so far and where you have difficulties doing these questions.Other problems...
$\displaystyle log_7 \frac{3}{147}$
$\displaystyle log_2(log_4 2 + log_4 8)^3$
$\displaystyle 3(4log_4 12-2log_4 9)$
thanks!~
Please show your work.
[quote=desiderius1;493586]more log questions:
$\displaystyle log_6 4 - log_6 3 - log_6 8$
so far I've done:
$\displaystyle log_6 4 - log_6 8 = log_6 {\frac48} = log_6 {\frac12}$
$\displaystyle log_6{\frac12} - log_6 3$
Stuck here.
Other problems...
$\displaystyle log_7 \frac{3}{147}$
$\displaystyle log_2(log_4 2 + log_4 8)^3$
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