Question

If \[ 4^x-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2x-1}, \] then \[ x= \]

• A. \(\frac{5}{2}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{3}{2}\)
• D. \(\frac{7}{2}\)

Hint:

When an equation contains powers with the same bases, first rewrite terms using common bases and then factorize similar exponential terms.

Answer 1

The Correct Option is C

Step 1: Write the given equation.
We have \[ 4^x-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2x-1} \] Since \[ 4^x=2^{2x}, \] the equation becomes \[ 2^{2x}-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2x-1} \]

Step 2: Bring similar base terms together.
Move the powers of \(2\) to one side and powers of \(3\) to the other side: \[ 2^{2x}+2^{2x-1}=3^{x+\frac{1}{2}}+3^{x-\frac{1}{2}} \]

Step 3: Factorize the left side.
\[ 2^{2x}+2^{2x-1}=2^{2x-1}(2+1) \] \[ =3\cdot 2^{2x-1} \]

Step 4: Factorize the right side.
\[ 3^{x+\frac{1}{2}}+3^{x-\frac{1}{2}} = 3^{x-\frac{1}{2}}(3+1) \] \[ =4\cdot 3^{x-\frac{1}{2}} \]

Step 5: Equate and simplify.
So, \[ 3\cdot 2^{2x-1}=4\cdot 3^{x-\frac{1}{2}} \] Now check option \(\frac{3}{2}\).
For \[ x=\frac{3}{2}, \] left side becomes \[ 3\cdot 2^{2\cdot \frac{3}{2}-1} = 3\cdot 2^2 = 12 \] Right side becomes \[ 4\cdot 3^{\frac{3}{2}-\frac{1}{2}} = 4\cdot 3^1 = 12 \] Both sides are equal.

Step 6: Final conclusion.
Therefore, \[ x=\frac{3}{2} \] Hence, \[ \boxed{\frac{3}{2}} \]