Messing with the bases of logarithms

**3k1yp2** · November 19th 2009, 05:55 PM

if log base a of x =c , and log base b of x = d, then what is the general statement that expresses log base (ab) of x in terms of c and d? plz help.

**lvleph** · November 19th 2009, 06:04 PM

If the problem state something like
$\log_a x = c$
$\log_b x = d$
$\log_{ab} x = ?$

If so, you will want to use a change of base rule here, i.e.
$\frac{\log_{a} x}{\log_{ab} a} = \log_{ab} x$

**Chris L T521** · November 19th 2009, 06:10 PM

Originally Posted by 3k1yp2

if log base a of x =c , and log base b of x = d, then what is the general statement that expresses log base (ab) of x in terms of c and d? plz help.

Use change of base formula:

$\log_a x=c\implies c=\frac{\ln x}{\ln a}\implies \ln a=\frac{\ln x}{c}$ and $\log_b x=d\implies d=\frac{\ln x}{\ln b}\implies \ln b=\frac{\ln x}{d}$ .

Now, $\log_{ab}x=\frac{\ln x}{\ln\!\left(ab\right)}=\frac{\ln x}{\ln a+\ln b}=\frac{\ln x}{\frac{1}{c}\ln x+\frac{1}{d}\ln x}=\frac{1}{\frac{c+d}{cd}}=\frac{cd}{c+d}$

Does this make sense?

**3k1yp2** · November 19th 2009, 06:14 PM

i do need to know what log base ab of x equals, but it has to be in terms of c and d.
i got that
log base a of x = c
log base b of x = d
so a^c = b^d , but im stuck there.

**3k1yp2** · November 19th 2009, 06:26 PM

that was extremely helpful, but i did get lost at the part where
ln(x) was ÷ by the sum of (1/c ln(x) ) and (1/d ln(x) ). where or how did you get those expressions in the denominator?

Nevermind, now i get it!
thank youuu! >singing<

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