Question

If a^p=b^q=c^r and b^2=ac, then prove that q= 2rp/r+p. whoever answers this question is a true genius. just get me the answer

Answer 1

VERY EASY BRO, Given: a^p=b^q=c^r, b^2=ac. To prove: q=2rp/r+p. Proof: let a^p= k^1/p (as when x^2 = 4, x= root of 4=2)as a^p= b^q= c^r, so, k^1/p= k^1/q= k^1/r. Also, b^2=ac so, (k^1/q)^2 = k^1/p * k^1/r. so, k^2/q = k^ r+p/pr (adding 1/p and 1/r). As, all the terms are equal(given), we can compare the powers, so, 2/q= r+p/pr Transposing the terms, we get, 2(rp)/r+p= q, and hence proved.

Answer 2

Answer:  The proof is doe below.

Step-by-step explanation:  We are given the following relations :

$a^p=b^q=c^r~~~\textup{and}~~~b^2=ac.$

We are to prove that :

$q=\dfrac{2rp}{r+p}.$

We will be using the following properties of exponents :

$(i)~a^b\times a^c=a^{b+c},\\\\(ii)~a^b=a^c~~~~~\Rightarrow b=c.$

Let us consider that

$a^p=b^q=c^r=k\\\\\Rightarrow a=k^\frac{1}{p},~~b=k^\frac{1}{q},~~c=k^\frac{1}{r}.$

Now, we have

$b^2=ac\\\\\Rightarrow (k^\frac{1}{q})^2=k^\frac{1}{p}\times k^\frac{1}{r}\\\\\Rightarrow k^\frac{2}{q}=k^{\frac{1}{p}+\frac{1}{r}}\\\\\Rightarrow \dfrac{2}{q}=\dfrac{1}{p}+\dfrac{1}{r}\\\\\\\Rightarrow \dfrac{2}{q}=\dfrac{r+p}{rp}\\\\\\\Rightarrow \dfrac{q}{2}=\dfrac{rp}{r+p}\\\\\\\Rightarrow q=\dfrac{2rp}{r+p}.$

Hence proved.