If $\displaystyle a^1^/^x = b^1^/^y = c^1^/^z$ & $\displaystyle b^2 = ac$, how do I find the value of $\displaystyle (x + z)/2y$?
Thanks,
Ron
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If $\displaystyle a^1^/^x = b^1^/^y = c^1^/^z$ & $\displaystyle b^2 = ac$, how do I find the value of $\displaystyle (x + z)/2y$?
Thanks,
Ron
take logs,Quote:
Originally Posted by rn5a
$\displaystyle \frac{\lg a}{x}=\frac{\lg b}{y}=\frac{\lg c}{z}$
substitute $\displaystyle x,z$ with $\displaystyle y$
Apply $\displaystyle \ln=$ neperian logarithm: $\displaystyle \displaystyle{a^1^/^x = b^1^/^y = c^1^/^z\Longrightarrow \frac{1}{x}\ln a=\frac{1}{y}\ln b=\frac{1}{z}\ln c$ , and etc.Quote:
Originally Posted by rn5a
Tonio
I don't have any idea about logs. How do I do it without using logs?
Thanks,
Ron
Hello, Ron!
Without logs? . . . This takes some gymnastics.
Quote:
$\displaystyle \text{If }a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}}\:\text{ and }\:b^2 = ac,\:\text{find the value of: }\:\dfrac{x + z}{2y}$
$\displaystyle \text{We have: }\:b^{\frac{1}{y}} \:=\:a^{\frac{1}{x}}$
$\displaystyle \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]
$\displaystyle \text{We have: }\:b^{\frac{1}{y}} \:=\:c^{\frac{1}{z}}$
$\displaystyle \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]: .$\displaystyle b^2 \;=\;a^{\frac{y}{x}}\!\cdot\!c^{\frac{z}{x}} $
We are told that: .$\displaystyle b^2 \,=\,a\!\cdot\!c$
$\displaystyle \text{Hence: }\;a^{\frac{y}{x}}\!\cdot\!c^{\frac{z}{x}} \:=\:a^1\!\cdot\!c^1 \quad\Rightarrow\quad \begin{Bmatrix}\frac{y}{x} \:=\:1 \\ \\[-4mm] \frac{z}{x} \:=\:1 \end{Bmatrix} \quad\Rightarrow\quad \begin{Bmatrix} y \:=\:x \\ z\:=\:x \end{Bmatrix}$
. . And we have: .$\displaystyle x = y = z$
$\displaystyle \displaystyle \text{Therefore: }\:\frac{x+z}{2y} \;=\;\frac{x+x}{2x} \;=\;\frac{2x}{2x} \;=\;1$
