Question

1. $(2^{2x + 2})(2^{3x - 7}) = 2^{25}$
2. $\frac{8^{\frac{x}{2}}}{4^{\frac{x}{3}}} = 2^{-\frac{5}{2}}$
3. $3 = (5^{\frac{1}{3}})(x^{\frac{1}{3}})$
4. $36^{2x - 7} = 6^{x - 5}$

Explanation:

Problem 6: \((2^{2x + 2})(2^{3x - 7}) = 2^{25}\)

Step 1: Use exponent rule \(a^m \cdot a^n = a^{m + n}\)

When multiplying two exponential terms with the same base, we add the exponents. So, \((2^{2x + 2})(2^{3x - 7}) = 2^{(2x + 2)+(3x - 7)}\)
Simplify the exponent: \((2x + 2)+(3x - 7)=2x + 2+3x - 7 = 5x - 5\)
So the equation becomes \(2^{5x - 5}=2^{25}\)

Step 2: Set exponents equal (since bases are equal)

If \(a^m=a^n\) and \(a>0,a
eq1\), then \(m = n\). Here \(a = 2\), so \(5x-5=25\)

Step 3: Solve for \(x\)

Add 5 to both sides: \(5x=25 + 5=30\)
Divide both sides by 5: \(x=\frac{30}{5}=6\)

Step 1: Rewrite 8 and 4 as powers of 2

We know that \(8 = 2^3\) and \(4=2^2\). So, \(8^{\frac{x}{2}}=(2^3)^{\frac{x}{2}}\) and \(4^{\frac{x}{3}}=(2^2)^{\frac{x}{3}}\)

Step 2: Use exponent rule \((a^m)^n=a^{mn}\)

For \((2^3)^{\frac{x}{2}}\), we have \(2^{3\times\frac{x}{2}}=2^{\frac{3x}{2}}\)
For \((2^2)^{\frac{x}{3}}\), we have \(2^{2\times\frac{x}{3}}=2^{\frac{2x}{3}}\)
The left - hand side becomes \(\frac{2^{\frac{3x}{2}}}{2^{\frac{2x}{3}}}\)

Step 3: Use exponent rule \(\frac{a^m}{a^n}=a^{m - n}\)

\(\frac{2^{\frac{3x}{2}}}{2^{\frac{2x}{3}}}=2^{\frac{3x}{2}-\frac{2x}{3}}\)
Find a common denominator for the exponents: \(\frac{3x}{2}-\frac{2x}{3}=\frac{9x - 4x}{6}=\frac{5x}{6}\)
So the equation is \(2^{\frac{5x}{6}}=2^{-\frac{5}{2}}\)

Step 4: Set exponents equal

Since the bases are equal (\(a = 2\)), we have \(\frac{5x}{6}=-\frac{5}{2}\)

Step 5: Solve for \(x\)

Multiply both sides by 6: \(5x=-\frac{5}{2}\times6=- 15\)
Divide both sides by 5: \(x=-3\)

Step 1: Use the property \(a^m\times b^m=(ab)^m\)

We can rewrite the right - hand side as \((5x)^{\frac{1}{3}}\) (since \(a^{\frac{1}{3}}\times b^{\frac{1}{3}}=(ab)^{\frac{1}{3}}\))
So the equation is \(3=(5x)^{\frac{1}{3}}\)

Step 2: Cube both sides

To eliminate the cube root, we cube both sides of the equation. \((3)^3=((5x)^{\frac{1}{3}})^3\)
We know that \((a^m)^n=a^{mn}\), so \(((5x)^{\frac{1}{3}})^3=(5x)^{\frac{1}{3}\times3}=5x\) and \(3^3 = 27\)
The equation becomes \(27 = 5x\)

Step 3: Solve for \(x\)

Divide both sides by 5: \(x=\frac{27}{5}=5.4\)

Answer:

\(x = 6\)

Problem 7: \(\frac{8^{\frac{x}{2}}}{4^{\frac{x}{3}}}=2^{-\frac{5}{2}}\)