Question

hw 9.1_log properties 1
score: 23.86/33 answered: 8/12
question 9
(a) rewrite the equation 6^w = 0.101 as a logarithm and solve to two decimals:
w =
(b) rewrite the equation 5^w = 4.06 as a logarithm and solve to two decimals:
w =
(c) rewrite the equation 3^w = 42.4 as a logarithm and solve to two decimals:
w =
question help: video

Explanation:

Step1: Recall the logarithm definition

For \(a^x = b\), we can rewrite it as \(x=\log_a b=\frac{\ln b}{\ln a}\) (change - of - base formula).

Part (a)

Step2: Rewrite the exponential equation as a logarithm

Given \(6^w = 0.101\), by the definition of logarithms, we can rewrite it as \(w=\log_6(0.101)\).

Step3: Use the change - of - base formula to calculate the value

Using the change - of - base formula \(\log_a b=\frac{\ln b}{\ln a}\), we have \(w = \frac{\ln(0.101)}{\ln(6)}\).
We know that \(\ln(0.101)\approx - 2.2923\) and \(\ln(6)\approx1.7918\).
Then \(w=\frac{-2.2923}{1.7918}\approx - 1.28\).

Part (b)

Step4: Rewrite the exponential equation as a logarithm

Given \(5^w = 4.06\), by the definition of logarithms, we can rewrite it as \(w=\log_5(4.06)\).

Step5: Use the change - of - base formula to calculate the value

Using the change - of - base formula \(\log_a b=\frac{\ln b}{\ln a}\), we have \(w=\frac{\ln(4.06)}{\ln(5)}\).
We know that \(\ln(4.06)\approx1.4017\) and \(\ln(5)\approx1.6094\).
Then \(w=\frac{1.4017}{1.6094}\approx0.87\).

Part (c)

Step6: Rewrite the exponential equation as a logarithm

Given \(3^w = 42.4\), by the definition of logarithms, we can rewrite it as \(w=\log_3(42.4)\).

Step7: Use the change - of - base formula to calculate the value

Using the change - of - base formula \(\log_a b=\frac{\ln b}{\ln a}\), we have \(w=\frac{\ln(42.4)}{\ln(3)}\).
We know that \(\ln(42.4)\approx3.7447\) and \(\ln(3)\approx1.0986\).
Then \(w=\frac{3.7447}{1.0986}\approx3.41\).

Answer:

(a) \(w\approx - 1.28\)
(b) \(w\approx0.87\)
(c) \(w\approx3.41\)