# Exponents

• Feb 15th 2011, 06:25 PM
rn5a
Exponents
If $a^1^/^x = b^1^/^y = c^1^/^z$ & $b^2 = ac$, how do I find the value of $(x + z)/2y$?

Thanks,

Ron
• Feb 15th 2011, 07:21 PM
Quote:

Originally Posted by rn5a
If $a^1^/^x = b^1^/^y = c^1^/^z$

take logs,

$\frac{\lg a}{x}=\frac{\lg b}{y}=\frac{\lg c}{z}$

substitute $x,z$ with $y$
• Feb 15th 2011, 07:22 PM
tonio
Quote:

Originally Posted by rn5a
If $a^1^/^x = b^1^/^y = c^1^/^z$ & $b^2 = ac$, how do I find the value of $(x + z)/2y$?

Thanks,

Ron

Apply $\ln=$ neperian logarithm: $\displaystyle{a^1^/^x = b^1^/^y = c^1^/^z\Longrightarrow \frac{1}{x}\ln a=\frac{1}{y}\ln b=\frac{1}{z}\ln c$ , and etc.

Tonio
• Feb 16th 2011, 06:41 AM
rn5a
I don't have any idea about logs. How do I do it without using logs?

Thanks,

Ron
• Feb 16th 2011, 07:23 AM
Soroban
Hello, Ron!

Without logs? . . . This takes some gymnastics.

Quote:

$\text{If }a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}}\:\text{ and }\:b^2 = ac,\:\text{find the value of: }\:\dfrac{x + z}{2y}$

$\text{We have: }\:b^{\frac{1}{y}} \:=\:a^{\frac{1}{x}}$
$\text{Raise both sides to the power }y\!:\;(b^{\frac{1}{y}})^y \;=\;(a^{\frac{1}{x}})^y \quad\Rightarrow\quad b \;=\;a^{\frac{y}{x}}$ .[1]

$\text{We have: }\:b^{\frac{1}{y}} \:=\:c^{\frac{1}{z}}$
$\text{Raise both sides to the power }y\!:\;(b^{\frac{1}{y}})^y \;=\;(c^{\frac{1}{z}})^y \quad\Rightarrow\quad b \;=\;c^{\frac{y}{z}}$ .[2]

Multiply [1] and [2]: . $b^2 \;=\;a^{\frac{y}{x}}\!\cdot\!c^{\frac{z}{x}}$

We are told that: . $b^2 \,=\,a\!\cdot\!c$

$\text{Hence: }\;a^{\frac{y}{x}}\!\cdot\!c^{\frac{z}{x}} \:=\:a^1\!\cdot\!c^1 \quad\Rightarrow\quad \begin{Bmatrix}\frac{y}{x} \:=\:1 \\ \\[-4mm] \frac{z}{x} \:=\:1 \end{Bmatrix} \quad\Rightarrow\quad \begin{Bmatrix} y \:=\:x \\ z\:=\:x \end{Bmatrix}$

. . And we have: . $x = y = z$

$\displaystyle \text{Therefore: }\:\frac{x+z}{2y} \;=\;\frac{x+x}{2x} \;=\;\frac{2x}{2x} \;=\;1$