1. ## logarithm bases assignment

hello everyone,

this is my first time on here, so i hope i'm doing it right

we have a year 12 maths assignment due in about a week, and it says:

Let log(a)x=c and log(b)x=d. Find the general statement that expresses log(ab)x in terms of c and d.

if it is.. how on earth is it right? lol.. how can i prove it?

thank you so much

2. Originally Posted by Missy
hello everyone,

this is my first time on here, so i hope i'm doing it right

we have a year 12 maths assignment due in about a week, and it says:

Let log(a)x=c and log(b)x=d. Find the general statement that expresses log(ab)x in terms of c and d.

if it is.. how on earth is it right? lol.. how can i prove it?

thank you so much

Welcome to the Forum

Formula

$\displaystyle log_N(M) = \frac{1}{log_{M}(N)}$

--------------------------------------------------

$\displaystyle log_x(a)+ log_x( b) = log_x(ab)$

$\displaystyle \frac{1}{log_a(x)}+\frac{1}{log_b(x)} = \frac{1}{log_{ab}(x)}$

$\displaystyle \frac{1}{c}+\frac{1}{d} = \frac{1}{log_{ab}(x)}$

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

3. ahh thank you thank you!
umm.. it depends if its good enough.. my teacher will pick the best out of four assignments..
thank you!!!

4. Hello, Missy!

Welcome aboard!

Here's another approach . . . It's considerably longer,
. . but it will impress/surprise/terrify your teacher.

Let $\displaystyle \log_aX \:=\:c$ and $\displaystyle \log_bX\:=\:d$
Find the general statement that expresses $\displaystyle \log_{ab}X$ in terms of $\displaystyle c$ and $\displaystyle d$.

is the answer: .$\displaystyle \log_{ab}X\:=\:\frac{cd}{c+d}$ ?

We have: .$\displaystyle \begin{array}{ccccccccccc}\log_aX \:=\:c &&\Rightarrow&& a^c \:=\:X &&\Rightarrow&& a \:=\:X^{\frac{1}{c}} & [1]\\ \log_bX \:=\:d && \Rightarrow& & b^d \:=\:X &&\Rightarrow&& b \:=\:X^{\frac{1}{d}} & [2] \end{array}$

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

Raise both sides to the power $\displaystyle \frac{cd}{c+d}$

. . $\displaystyle (ab)^{\frac{cd}{c+d}} \;=\;\left(X^{\frac{c+d}{cd}}\right)^{\frac{cd}{c+ d}} \quad\Rightarrow\quad (ab)^{\frac{cd}{c+d}} \;=\;X$

Take logs, base $\displaystyle ab$

. . $\displaystyle \log_{ab}\left[(ab)^{\frac{cd}{c+d}}\right] \;=\;\log_{ab}(X) \quad\Rightarrow\quad \log_{ab}(X) \;=\;\frac{cd}{c+d}\cdot\underbrace{\log_{ab}(ab)} _{\text{This is 1}}$

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

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

[/quote]
i agree with u, but the thing is, i dont understand how u can substitute values for a, b, and x .(for testing the formula)
soo can someone plz help me, cuz i am having this problem.
and can u do the subtitution on the same formula!!!! :P

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

thnx

plzzz help me i need this for tomorrow