Question

2^{2+x}*3^x= 2/3*6^{3x-1}

Answer 1

To solve the equation \(2^{2+x} \cdot 3^x = \frac{2}{3} \cdot 6^{3x-1},\) you can start by simplifying both sides.

First, notice that \(6\) can be expressed as \(2 \cdot 3.\) Replace \(6\) with \(2 \cdot 3\) in the equation:

\[2^{2+x} \cdot 3^x = \frac{2}{3} \cdot (2 \cdot 3)^{3x-1}.\]

Now, simplify further. Expand the powers:

\[2^{2+x} \cdot 3^x = \frac{2}{3} \cdot 2^{3(3x-1)} \cdot 3^{3x-1}.\]

Next, combine like terms:

\[2^{2+x} \cdot 3^x = \frac{2}{3} \cdot 2^{9x-3} \cdot 3^{3x-1}.\]

Now, since the bases are the same on both sides of the equation, you can equate the exponents:

\[2+x = 9x-3.\]

Solve for \(x\):

\[x = \frac{1}{4}.\]

So, \(x = \frac{1}{4}\) is the solution to the equation.

Answer 2

x=1

Explanation:

Notice that 6=2*3, so we can express 6^(3x-1) as {2^(3x-1)}*{3^(3x-1)}.

Now, substitute this into the equation:

2^{2+x}*3^x= 2/3*{2^(3x-1)}*{3^(3x-1)}

To simplify, let's clear the fraction by multiplying both sides by 3:

3*2^{2+x}*3^x= 3*(2/3)*{2^(3x-1)}*{3^(3x-1)}

3*2^{2+x}*3^x= 2*{2^(3x-1)}*{3^(3x-1)}

Now, use the properties of exponents. Distribute the exponents on both sides:

3*2^2*2^x*3^x= 2*2^3x*2^(-1)*3^3x*3^(-1)

Simplify both sides:

12*2^x*3^x=(2/6)*2^3x*3^3x

12*6^x=(6^3x)/3

6^3x=36*6^x

(6^x)^3=36*6^x

36=(6^x)^2

6^2=(6^x)^2

6=6^x

6^1=6^x

Therefore:

x=1