1. ## Proving

Show that if
logb(a)=c and
logy(b)=c, then
loga(y)=c^-2

2. Hello KingV15
Originally Posted by KingV15
Show that if
logb(a)=c and
logy(b)=c, then
loga(y)=c^-2
I generally reckon that the easiest way to solve log problems is to get rid of the logs as soon as possible! So let's try that here.

$\displaystyle \log_b(a)=c$
$\displaystyle \Rightarrow a=b^c$ ...(1)
$\displaystyle \log_y(b)=c$
$\displaystyle \Rightarrow b=y^c$ ...(2)
and we need to show that:
$\displaystyle \log_a(y)=c^{-2}$
in other words that:
$\displaystyle y = a^{(c^{-2})}$
Now from (1):
$\displaystyle b = a^{(c^{-1})}$ ...(3)
And from (2):
$\displaystyle y=b^{(c^{-1})}$
$\displaystyle = \big(a^{(c^{-1}}\big)^{(c^{-1})}$ from (3)

$\displaystyle =a^{(c^{-2})}$
$\displaystyle \Rightarrow \log_a(y)=c^{-2}$

3. Hello, KingV15!

You used my favorite approach.

Here's another . . .

Show that if: .$\displaystyle \begin{array}{ccc}\log_b(a)\:=\:c & [1] \\ \log_y(b)\:=\:c & [2]\end{array}$

. . then: .$\displaystyle \log_a(y)\:=\:c^{-2}$

Base-change Formula: .$\displaystyle \log_b(N) \:=\:\frac{\log(N)}{\log(b)}$

$\displaystyle \begin{array}{ccccccc}\text{From [1], we have:} & \dfrac{\log(a)}{\log(b)} \:=\:c & \Rightarrow & \dfrac{\log(b)}{\log(a)} \:=\:\dfrac{1}{c} & [3] \\ \\[-3mm] \text{From [2], we have:} & \dfrac{\log(b)}{\log(y)} \:=\:c & \Rightarrow & \dfrac{\log(y)}{\log(b)} \:=\:\dfrac{1}{c} & [4] \end{array}$

Multiply [3] by [4]: .$\displaystyle \frac{\log(b)}{\log(a)}\cdot\frac{\log(y)}{\log(b) } \:=\:\frac{1}{c}\cdot\frac{1}{c} \quad\Rightarrow\quad \frac{\log(y)}{\log(a)} \:=\:\frac{1}{c^2}$

. . Therefore: .$\displaystyle \log_a(y) \:=\:c^{-2}$