If a^p = b^q = c^r = abc, then pqr = ? Option 1: (qp + qr + pr)^2 Option 2: (p + q + r)^2 Option 3: pq(qr + rp) Option 4: pq + qr + pr

Answer: pq + qr + pr. Solution: Since a^p = b^q = c^r = abc, let all of them equal k. a^p = k a = k^(1/p) b^q = k b = k^(1/q) c^r = k c = k^(1/r) abc = k^(1/p) * k^(1/q) * k^(1/r) abc = k^(1/p + 1/q + 1/r) Since abc = k = k^1, comparing the exponents: 1/p + 1/q + 1/r = 1 Taking pqr as the common denominator: (qr + pr + pq)/pqr = 1 pqr = pq + qr + pr

Algebra

Indices

Medium

If $a^p = b^q = c^r = abc$ , then $pqr = ?$

$(qp + qr + pr)^2$

$(p + q + r)^2$

$pq(qr + rp)$

$pq + qr + pr$

✅ Correct Option: 4

Since $a^p = b^q = c^r = abc$ , let all of them equal $k$ .

$a^p = k \Rightarrow a = k^{1/p}$

$b^q = k \Rightarrow b = k^{1/q}$

$c^r = k \Rightarrow c = k^{1/r}$

$abc = k^{1/p} \cdot k^{1/q} \cdot k^{1/r}$

$abc = k^{\frac{1}{p} + \frac{1}{q} + \frac{1}{r}}$

Since $abc = k = k^1$ , comparing the exponents:

$\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r} = 1$

Taking $pqr$ as the common denominator:

$\dfrac{qr + pr + pq}{pqr} = 1$

$pqr = pq + qr + pr$

BMSAT Kozhikode 2025

If $x^{\frac{1}{12}}=49^{\frac{1}{24}}$ , the value of $x$ is:

BMSAT Kozhikode 2026

If $2^x/(1 + 2^x) = 1/3$ , then $8^x/(1-8^x) = ?$

BMSAT Kozhikode 2026

If $a = \sqrt{11} - \sqrt{10}$ , $b = \sqrt{12} - \sqrt{11}$ , and $c = \sqrt{13} - \sqrt{12}$ , then which of the following is true?

BMSAT Kozhikode 2026

If $2^{n+m}/(2^{n-m}) = 16$ and $a = 2^{1/10}$ , then, $(a^{2m+n-p})^2/(a^{m-2n+2p})^{-1} = ?$

BMSAT Kozhikode 2026

The value of q that satisfies $3^p + 5^q = 52$ and $3^{p-1} + 5^{q-1} = 14$ will be

BMSAT Kozhikode 2026

If $\sqrt{2^n} = 1024$ , then $3^{(n/4 - 4)}$ is = ?

If ap=bq=cr=abca^p = b^q = c^r = abcap=bq=cr=abc, then pqr=?pqr = ?pqr=?

Solution

Related questions:

Keyboard Shortcuts

BMSAT Kozhikode Past Year Questions (Topic-Wise)

If $a^p = b^q = c^r = abc$ , then $pqr = ?$