Question

1. what is the solution to $5(2)^{19x}=50$? (1) $x = \frac{\log(50)}{19}$ (2) $x = \frac{\log_{2}(10)}{19}$ (3) $x = \frac{\log_{2}(45)}{19}$ (4) $x = \frac{5}{19}$ 11. which sequence has a common ratio of $\frac{1}{2}$? (1) $-\frac{1}{4}a, -\frac{1}{8}a, -\frac{1}{16}a, -\frac{1}{32}a,\dots$ (2) $\frac{1}{32}a, \frac{1}{16}a, \frac{1}{8}a, \frac{1}{4}a,\dots$ (3) $20a, \frac{39}{2}a, 19a, \frac{37}{2}a,\dots$ (4) $22a, 22.5a, 23a, 23.5a,\dots$ 12. given $m \

eq 0$ and $\left(17^{\frac{1}{m}}\
ight)^{n} = 17^{2}$, what is $n$ in terms of $m$? (1) $2m$ (2) $\frac{2}{m}$ (3) $\frac{m}{2}$ (4) $2^{m}$

Explanation:

Question 10

Step1: Isolate the exponential term

Divide both sides by 5:
$\frac{5(2)^{19x}}{5} = \frac{50}{5}$
$2^{19x} = 10$

Step2: Convert to logarithmic form

Use $\log_b(a)=c \iff b^c=a$ (base 2):
$19x = \log_2(10)$

Step3: Solve for x

Divide both sides by 19:
$x = \frac{\log_2(10)}{19}$

Step1: Recall common ratio definition

Common ratio $r = \frac{\text{next term}}{\text{current term}}$, target $r=\frac{1}{2}$.

Step2: Test Option (1)

$r = \frac{-\frac{1}{8}a}{-\frac{1}{4}a} = \frac{1}{2}$

Step3: Verify other options (elimination)

• Option (2): $r = \frac{\frac{1}{16}a}{\frac{1}{32}a}=2$ (not $\frac{1}{2}$)
• Option (3): $r = \frac{\frac{39}{2}a}{20a}=\frac{39}{40}$ (not $\frac{1}{2}$)
• Option (4): This is an arithmetic sequence (common difference $0.5a$, no common ratio)

Step1: Apply exponent power rule

Use $(b^p)^q = b^{pq}$:
$17^{\frac{1}{m} \cdot n} = 17^2$

Step2: Equate exponents (same base)

Since bases are equal, exponents are equal:
$\frac{n}{m} = 2$

Step3: Solve for n

Multiply both sides by $m$:
$n = 2m$

Answer:

(2) $x = \frac{\log_{2}(10)}{19}$

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Question 11