# Proving

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• Jan 13th 2010, 04:40 AM
KingV15
Proving
Show that if
logb(a)=c and
logy(b)=c, then
loga(y)=c^-2
• Jan 13th 2010, 05:33 AM
Grandad
Hello KingV15
Quote:

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.

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

$=a^{(c^{-2})}$
$\Rightarrow \log_a(y)=c^{-2}$
Grandad
• Jan 13th 2010, 10:15 AM
Soroban
Hello, KingV15!

Good job, Grandad!
You used my favorite approach.

Here's another . . .

Quote:

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

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

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

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

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

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