Question

If x^a = y^b = z^c and y^2 = xz; prove that b = (2ac)/(a + c)

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Answer 1

Hey there,

Here is Ur answer,

➡x^a=y^b=z^c=k(let)

➡x=k^(1/a)

➡y=k^(1/b)

➡z=k^(1/c)

Given,y^2=xz

➡k^[(1/b)2]=k^(1/a)×k^(1/c)

➡k^(2/b)=k^(1/a+1/c)

Taking log both sides with base k,we get

➡2/b=1/a+1/c

➡2/b=(a+c)/ac

➡✔✔b=2ac/(a+b)✔✔

Hope it helps

✌✌✌✌

Answer 2

Given: The expression, $x^a=y^b=z^c$ and $y^2=xz$

To prove: $b=\frac{2ac}{a+c}$

Step-by-step explanation:

We are given that $x^a=y^b=z^c$,

Let's take a constant, 'k', which is equal to the expression,

So, we get:

$x^a=y^b=z^c=k$

Now, taking $x^a$, $y^b$ and $z^c$ one by one, we get:

For the value of x:

$x^a=k\\x=k^\frac{1}{a}$                      ...(i)

For the value of y:

$y^b=k\\y=k^\frac{1}{b}$                      ...(ii)

For the value of z:

$z^c=k\\z=k^\frac{1}{c}$                      ...(iii)

Also, we are given the condition that: $y^2=xz$

Now, substituting (i), (ii) and (iii) in this condition, we get:

$k^\frac{1}{b}^{(2)}=k^\frac{1}{a}\times k^\frac{1}{c}$

Taking logarithm both sides, we get:

$log|k^\frac{1}{b}^{(2)}|=log|k^\frac{1}{a}\times k^\frac{1}{c}|$

Using the properties of logarithm:

$log|x^y|=ylog|x|$   and  $log|xy|=log|x|+log|y|$, we get:

$\frac{2}{b}log|k|=log|k^\frac{1}{a}|+log|k^\frac{1}{c}|$

Again using the properties of logarithm, we get:

$\frac{2}{b}log|k|=\frac{1}{a}log|k|+\frac{1}{c}log|k|$

We can cancel out the logarithmic terms on both sides and then we get:

$\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$

$\frac{2}{b}=\frac{a+c}{ac}$

Using cross mulitplication, we get:

$2ac=b(a+c)$

Or we can say:

$b=\frac{2ac}{a+c}$

Hence proved.