Here is a quetion I couldn't solve.

Let loga (x) = c and logb (x) = d.

Find the general statement that expresses logab (x), in terms of c and d.

I would appreciate your help, thanks a lot,

Amine :)

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- Apr 17th 2009, 02:13 AMAminekhadirLogarithms and patterns
Here is a quetion I couldn't solve.

Let loga (x) = c and logb (x) = d.

Find the general statement that expresses logab (x), in terms of c and d.

I would appreciate your help, thanks a lot,

Amine :) - Apr 17th 2009, 03:13 AMrubix
loga (x) = c

logb (x) = d

logab (x) = ?

a^c = x and b^d = x

a = x ^ (1/c) and b = x ^ (1/d)

ab = x^ 1/c x ^ 1/d

= x ^ (d+c/cd)

logab (x) = d+c / cd

humm i might have flipped something (Wondering) - Apr 17th 2009, 03:55 AMSoroban
Hello, rubix!

Ya done good!

I'll put it in LaTex . . .

Quote:

$\displaystyle \begin{array}{c}\log_a (x) \:=\: c \\ \log_b(x) \:=\:d\end{array}\qquad \log_{ab}(x) \:=\: ?$

$\displaystyle \begin{array}{ccccc}a^c \:= \:x & \Rightarrow & a \:=\:x^{\frac{1}{x}} & [1] \\ b^d \:=\: x & \Rightarrow & b \:=\:x^{\frac{1}{d}} & [2]\end{array}$

Multiply [1] and [2]: .$\displaystyle ab \:=\:x^{\frac{1}{c}}\cdot x^{\frac{1}{d}} \;=\;x^{\frac{d+c}{cd}}$

Therefore: .$\displaystyle \log_{ab}(x) \:=\: \frac{c+d}{cd}$

Good work!

- Apr 17th 2009, 04:11 AMrubix
thnx for conforming

**Soroban**...and thnx for nice well formatted answe.

cheers. - Apr 17th 2009, 12:21 PMAminekhadir
Thanks a lot for your help! :)