logarithms rules

**eq123** · February 19th 2010, 06:25 AM

$\frac {2}{log_{5} 3} = 2 \log_{3} 5 = 2 \frac{\log 5}{\log 3}$

what rules have been used in these steps?

**HallsofIvy** · February 19th 2010, 06:56 AM

Originally Posted by eq123

$\frac {2}{log_{5} 3} = 2 \log_{3} 5 = 2 \frac{\log 5}{\log 3}$

what rules have been used in these steps?

In addition to the usual "rules of logarithms", you need the definition of logarithm.

If $x= log_3 5$ , then $3^x= 5$ . Now take the logarithm, base 5, of that: $log_5 3^x= log_5 5$ and use the "law of logarithms" $log(a^b)= b log(a)$ to get [tex]xlog_5(3)= 1[tex] and so [tex]x= \frac{1}{log_5(3)}.

For the last take log of both sides of $3^x= 5$ : $log(3^x)= x log(3)= log(5)$ and divide both sides of the equation by log(3).

The "2" is just there to confuse things. Once you have those three different forms of "x" equal, multiply each by 2.

**Soroban** · February 19th 2010, 07:11 AM

Hello, eq123!

. . . $\begin{array}{ccccccc} [1] && [2] && [3] \\ \\[-3mm] \dfrac {2}{\log_5 3} & =& 2 \log_3 5 &=& 2\,\dfrac{\log 5}{\log 3} \end{array}$

What rules have been used in these steps?

Going from [1] to [2], the Reciprocal Rule is used: . $\log_b(a) \:=\:\frac{1}{\log_a(b)}$

Going from [2] to [3], the Base-Change Formula: . $\log_b(a) \:=\:\frac{\log(a)}{\log(b)}$

**eq123** · February 19th 2010, 07:20 AM

thank you for your help.. it's all clear now..

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