Question

1. $49^{\frac{x}{4}-1} = 343^{\frac{x}{3}}$ 39. $2 = \left(4^{\frac{1}{3}}\

ight)\left(2^{\frac{x}{3}}\
ight)$

Explanation:

Problem 37:

Step1: Express bases as powers of 7

$49 = 7^2$ and $343 = 7^3$, so rewrite the equation:
$(7^2)^{\frac{x}{4}-1} = (7^3)^{\frac{x}{3}}$

Step2: Apply exponent rule $(a^m)^n = a^{mn}$

Left side: $7^{2(\frac{x}{4}-1)} = 7^{\frac{x}{2}-2}$
Right side: $7^{3(\frac{x}{3})} = 7^x$

Step3: Set exponents equal (since bases are equal)

$\frac{x}{2}-2 = x$

Step4: Solve for x

Subtract $\frac{x}{2}$ from both sides: $-2 = x - \frac{x}{2} = \frac{x}{2}$
Multiply both sides by 2: $x = -4$

Problem 39:

Step1: Express 4 as a power of 2

$4 = 2^2$, so rewrite $4^{\frac{1}{3}}$:
$2 = (2^2)^{\frac{1}{3}} \cdot 2^{\frac{x}{3}}$

Step2: Apply exponent rule $(a^m)^n = a^{mn}$

$(2^2)^{\frac{1}{3}} = 2^{\frac{2}{3}}$, so the equation becomes:
$2 = 2^{\frac{2}{3}} \cdot 2^{\frac{x}{3}}$

Step3: Apply exponent rule $a^m \cdot a^n = a^{m+n}$

Right side: $2^{\frac{2}{3} + \frac{x}{3}} = 2^{\frac{x+2}{3}}$

Step4: Set exponents equal (since bases are equal)

$1 = \frac{x+2}{3}$ (because $2 = 2^1$)

Step5: Solve for x

Multiply both sides by 3: $3 = x + 2$
Subtract 2: $x = 1$

Answer:

s:
Problem 37: $\boldsymbol{x = -4}$
Problem 39: $\boldsymbol{x = 1}$