logarithms

**requal** · February 22nd 2009, 01:48 AM

Evaluate $\log_{9} 49-\log_{3} 7$

**Moo** · February 22nd 2009, 01:59 AM

Hello,

Originally Posted by requal

Evaluate $\log_{9} 49-\log_{3} 7$

Recall these formulae :
$\log_a(b)=\frac{\ln(b)}{\ln(a)}$
$\log(a^b)=b \log(a)$

So $\log_9 49=\frac{\ln(49)}{\ln(9)}=\frac{\ln(7^2)}{\ln(3^2) }=\frac{\ln(7)}{\ln(3)}=\log_3(7)$

Thus...

**Soroban** · February 22nd 2009, 05:13 AM

Hello, requal!

Evaluate: . $\log_{9} 49-\log_{3} 7$

$\text{Let: }\,\log_949 \:=\:p$ .[1]

Then we have: . $9^p \:=\:49 \quad\Rightarrow\quad (3^2)^p \:=\:7^2 \quad\Rightarrow\quad 3^{2p} \:=\:7^2$

Take the square root of both sides: . $\left(3^{2p}\right)^{\frac{1}{2}} \:=\:\left(7^2\right)^{\frac{1}{2}} \quad\Rightarrow\quad 3^p \:=\:7$

This can be written: . $p \:=\:\log_37$ .[2]

From [1] and [2], we have: . $\log_949 \:=\:\log_37$

. . Therefore: . $\log_949 - \log37 \;=\;0$

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