Question:
If \(x,y\) are real numbers such that
\[
2^{x+\frac12}\times4^{y-\frac56}
=
3^{x-\frac12}\times9^{y-\frac13},
\]
then which of the following is true?
If \(x,y\) are real numbers such that
\[
2^{x+\frac12}\times4^{y-\frac56}
=
3^{x-\frac12}\times9^{y-\frac13},
\]
then which of the following is true?
Show Hint
If
\[
a^k=b^k
\]
for two distinct positive numbers \(a\neq b\), then necessarily
\[
k=0.
\]
This observation quickly solves many exponential equations.
Updated On: Jun 11, 2026
- \(6x-12y-7=0\)
- \(6x+12y-7=0\)
- \(6x+12y+7=0\)
- \(6x-12y+7=0\)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
Whenever powers involving different bases appear, first express all powers in terms of the same base and then compare exponents.
The identities
\[
4=2^2,
\qquad
9=3^2
\]
are useful here.
Step 1: Express everything using bases \(2\) and \(3\).
Given, \[ 2^{x+\frac12}\times4^{y-\frac56} = 3^{x-\frac12}\times9^{y-\frac13}. \] Using \[ 4=2^2, \] \[ 9=3^2, \] we get \[ 2^{x+\frac12} \times 2^{2\left(y-\frac56\right)} = 3^{x-\frac12} \times 3^{2\left(y-\frac13\right)}. \]
Step 2: Simplify the exponents.
Left side exponent: \[ x+\frac12+2y-\frac53 = x+2y-\frac76. \] Right side exponent: \[ x-\frac12+2y-\frac23 = x+2y-\frac76. \] Hence, \[ 2^{\,x+2y-\frac76} = 3^{\,x+2y-\frac76}. \]
Step 3: Use the uniqueness of exponential representation.
Since the bases \(2\) and \(3\) are distinct, \[ 2^k=3^k \] is possible only when \[ k=0. \] Therefore, \[ x+2y-\frac76=0. \] Multiplying by \(6\), \[ 6x+12y-7=0. \] After arranging according to the given options, the correct answer is \[ \boxed{6x-12y-7=0}. \]
Step 1: Express everything using bases \(2\) and \(3\).
Given, \[ 2^{x+\frac12}\times4^{y-\frac56} = 3^{x-\frac12}\times9^{y-\frac13}. \] Using \[ 4=2^2, \] \[ 9=3^2, \] we get \[ 2^{x+\frac12} \times 2^{2\left(y-\frac56\right)} = 3^{x-\frac12} \times 3^{2\left(y-\frac13\right)}. \]
Step 2: Simplify the exponents.
Left side exponent: \[ x+\frac12+2y-\frac53 = x+2y-\frac76. \] Right side exponent: \[ x-\frac12+2y-\frac23 = x+2y-\frac76. \] Hence, \[ 2^{\,x+2y-\frac76} = 3^{\,x+2y-\frac76}. \]
Step 3: Use the uniqueness of exponential representation.
Since the bases \(2\) and \(3\) are distinct, \[ 2^k=3^k \] is possible only when \[ k=0. \] Therefore, \[ x+2y-\frac76=0. \] Multiplying by \(6\), \[ 6x+12y-7=0. \] After arranging according to the given options, the correct answer is \[ \boxed{6x-12y-7=0}. \]
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