Question

1. $6 cdot 10^n = 18$
2. $-4 cdot 10^n = -25$
3. $1.6 cdot 10^x = 42$
4. $-5 cdot 10^y = -17$

Explanation:

Problem 17: $6 \cdot 10^p = 18$

Step1: Isolate the power term

Divide both sides by 6:
$\frac{6 \cdot 10^p}{6} = \frac{18}{6}$
$10^p = 3$

Step2: Solve for $p$ using log

Take base-10 log of both sides:
$\log_{10}(10^p) = \log_{10}(3)$
$p = \log_{10}(3) \approx 0.477$

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Problem 18: $-4 \cdot 10^q = -25$

Step1: Isolate the power term

Divide both sides by -4:
$\frac{-4 \cdot 10^q}{-4} = \frac{-25}{-4}$
$10^q = 6.25$

Step2: Solve for $q$ using log

Take base-10 log of both sides:
$\log_{10}(10^q) = \log_{10}(6.25)$
$q = \log_{10}(6.25) \approx 0.796$

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Problem 19: $1.6 \cdot 10^x = 42$

Step1: Isolate the power term

Divide both sides by 1.6:
$\frac{1.6 \cdot 10^x}{1.6} = \frac{42}{1.6}$
$10^x = 26.25$

Step2: Solve for $x$ using log

Take base-10 log of both sides:
$\log_{10}(10^x) = \log_{10}(26.25)$
$x = \log_{10}(26.25) \approx 1.419$

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Problem 20: $-5 \cdot 10^y = -17$

Step1: Isolate the power term

Divide both sides by -5:
$\frac{-5 \cdot 10^y}{-5} = \frac{-17}{-5}$
$10^y = 3.4$

Step2: Solve for $y$ using log

Take base-10 log of both sides:
$\log_{10}(10^y) = \log_{10}(3.4)$
$y = \log_{10}(3.4) \approx 0.531$

Answer:

1. $p = \log_{10}(3) \approx 0.477$
2. $q = \log_{10}(6.25) \approx 0.796$
3. $x = \log_{10}(26.25) \approx 1.419$
4. $y = \log_{10}(3.4) \approx 0.531$