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To find the distance [tex]\(d\)[/tex] that the tornado has traveled when the wind speed [tex]\(S\)[/tex] is 150 miles per hour, we start with the given equation:
[tex]\[ S = 93 \log_{10}(d) + 65 \][/tex]
We are given [tex]\( S = 150 \)[/tex]. Plugging it into the equation, we get:
[tex]\[ 150 = 93 \log_{10}(d) + 65 \][/tex]
First, we isolate the logarithmic term by subtracting 65 from both sides:
[tex]\[ 150 - 65 = 93 \log_{10}(d) \][/tex]
This simplifies to:
[tex]\[ 85 = 93 \log_{10}(d) \][/tex]
Next, we solve for [tex]\( \log_{10}(d) \)[/tex] by dividing both sides by 93:
[tex]\[ \log_{10}(d) = \frac{85}{93} \][/tex]
Evaluating the right-hand side, we find:
[tex]\[ \log_{10}(d) \approx 0.9139784946236559 \][/tex]
To solve for [tex]\(d\)[/tex], we need to exponentiate both sides with base 10:
[tex]\[ d = 10^{0.9139784946236559} \][/tex]
Calculating the value, we get:
[tex]\[ d \approx 8.203109232014354 \][/tex]
Finally, we round this distance to the nearest tenth:
[tex]\[ d \approx 8.2 \][/tex]
Thus, the tornado traveled approximately [tex]\( \boxed{8.2} \)[/tex] miles.
[tex]\[ S = 93 \log_{10}(d) + 65 \][/tex]
We are given [tex]\( S = 150 \)[/tex]. Plugging it into the equation, we get:
[tex]\[ 150 = 93 \log_{10}(d) + 65 \][/tex]
First, we isolate the logarithmic term by subtracting 65 from both sides:
[tex]\[ 150 - 65 = 93 \log_{10}(d) \][/tex]
This simplifies to:
[tex]\[ 85 = 93 \log_{10}(d) \][/tex]
Next, we solve for [tex]\( \log_{10}(d) \)[/tex] by dividing both sides by 93:
[tex]\[ \log_{10}(d) = \frac{85}{93} \][/tex]
Evaluating the right-hand side, we find:
[tex]\[ \log_{10}(d) \approx 0.9139784946236559 \][/tex]
To solve for [tex]\(d\)[/tex], we need to exponentiate both sides with base 10:
[tex]\[ d = 10^{0.9139784946236559} \][/tex]
Calculating the value, we get:
[tex]\[ d \approx 8.203109232014354 \][/tex]
Finally, we round this distance to the nearest tenth:
[tex]\[ d \approx 8.2 \][/tex]
Thus, the tornado traveled approximately [tex]\( \boxed{8.2} \)[/tex] miles.
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