Question

If
\[ 4^x-3^{x-\frac12}=3^{x+\frac12}-2^{2x-1}, \] then the value of \(x\) is

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

Hint:

In exponential equations, first rewrite all terms using common bases wherever possible, then group and factor similar exponential expressions.

Answer 1

The Correct Option is D

Step 1: Write the given equation.
Given equation is
\[ 4^x-3^{x-\frac12}=3^{x+\frac12}-2^{2x-1} \]
Since
\[ 4^x=2^{2x}, \] the equation becomes
\[ 2^{2x}-3^{x-\frac12}=3^{x+\frac12}-2^{2x-1} \]

Step 2: Bring similar exponential terms together.
Move the powers of \(2\) to one side and powers of \(3\) to the other side:
\[ 2^{2x}+2^{2x-1}=3^{x+\frac12}+3^{x-\frac12} \]

Step 3: Factor both sides.
On the left side,
\[ 2^{2x}+2^{2x-1}=2^{2x-1}(2+1) \]
\[ =3\cdot 2^{2x-1} \]
On the right side,
\[ 3^{x+\frac12}+3^{x-\frac12}=3^{x-\frac12}(3+1) \]
\[ =4\cdot 3^{x-\frac12} \]
Thus,
\[ 3\cdot 2^{2x-1}=4\cdot 3^{x-\frac12} \]

Step 4: Check the matching option.
Put
\[ x=\frac32 \]
Left side of original equation:
\[ 4^{\frac32}-3^{\frac32-\frac12} \]
\[ =8-3 \]
\[ =5 \]
Right side of original equation:
\[ 3^{\frac32+\frac12}-2^{2\cdot\frac32-1} \]
\[ =3^2-2^2 \]
\[ =9-4 \]
\[ =5 \]
Both sides are equal. Therefore,
\[ x=\frac32 \]

Step 5: Final conclusion.
Hence,
\[ \boxed{\frac32} \]