Answer :
To solve the equation [tex]\(4 \cdot 10^{-3x} = 18\)[/tex] for [tex]\(x\)[/tex], we should convert the equation into logarithmic form to make it easier to handle. Let's go through the steps:
1. Rewrite the equation using logarithms:
[tex]\[ 4 \cdot 10^{-3x} = 18 \][/tex]
2. Take the logarithm of both sides (using base 10):
[tex]\[ \log_{10}(4 \cdot 10^{-3x}) = \log_{10}(18) \][/tex]
3. Use the property of logarithms that [tex]\(\log_b(ab) = \log_b(a) + \log_b(b)\)[/tex]:
[tex]\[ \log_{10}(4) + \log_{10}(10^{-3x}) = \log_{10}(18) \][/tex]
4. Simplify the expression. Recall that [tex]\(\log_{10}(10^y) = y\)[/tex], so:
[tex]\[ \log_{10}(4) + (-3x) \log_{10}(10) = \log_{10}(18) \][/tex]
Since [tex]\(\log_{10}(10) = 1\)[/tex], the equation becomes:
[tex]\[ \log_{10}(4) - 3x = \log_{10}(18) \][/tex]
5. Isolate the term involving [tex]\(x\)[/tex]:
[tex]\[ -3x = \log_{10}(18) - \log_{10}(4) \][/tex]
6. Solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{\log_{10}(4) - \log_{10}(18)}{-3} \][/tex]
Now let's calculate the numerical values using the given logarithmic values:
1. [tex]\(\log_{10}(4) \approx 0.6021\)[/tex]
2. [tex]\(\log_{10}(18) \approx 1.2553\)[/tex]
Substitute these values into the equation:
[tex]\[ x = \frac{0.6021 - 1.2553}{-3} \][/tex]
Calculate the numerator:
[tex]\[ 0.6021 - 1.2553 \approx -0.6532 \][/tex]
Now divide by [tex]\(-3\)[/tex]:
[tex]\[ x = \frac{-0.6532}{-3} \approx 0.2177 \][/tex]
Therefore, the exact value of [tex]\(x\)[/tex] is:
[tex]\[ x = \frac{\log_{10}(4) - \log_{10}(18)}{-3} \][/tex]
And the approximate value of [tex]\(x\)[/tex], rounded to the nearest thousandth, is:
[tex]\[ x \approx 0.218 \][/tex]
1. Rewrite the equation using logarithms:
[tex]\[ 4 \cdot 10^{-3x} = 18 \][/tex]
2. Take the logarithm of both sides (using base 10):
[tex]\[ \log_{10}(4 \cdot 10^{-3x}) = \log_{10}(18) \][/tex]
3. Use the property of logarithms that [tex]\(\log_b(ab) = \log_b(a) + \log_b(b)\)[/tex]:
[tex]\[ \log_{10}(4) + \log_{10}(10^{-3x}) = \log_{10}(18) \][/tex]
4. Simplify the expression. Recall that [tex]\(\log_{10}(10^y) = y\)[/tex], so:
[tex]\[ \log_{10}(4) + (-3x) \log_{10}(10) = \log_{10}(18) \][/tex]
Since [tex]\(\log_{10}(10) = 1\)[/tex], the equation becomes:
[tex]\[ \log_{10}(4) - 3x = \log_{10}(18) \][/tex]
5. Isolate the term involving [tex]\(x\)[/tex]:
[tex]\[ -3x = \log_{10}(18) - \log_{10}(4) \][/tex]
6. Solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{\log_{10}(4) - \log_{10}(18)}{-3} \][/tex]
Now let's calculate the numerical values using the given logarithmic values:
1. [tex]\(\log_{10}(4) \approx 0.6021\)[/tex]
2. [tex]\(\log_{10}(18) \approx 1.2553\)[/tex]
Substitute these values into the equation:
[tex]\[ x = \frac{0.6021 - 1.2553}{-3} \][/tex]
Calculate the numerator:
[tex]\[ 0.6021 - 1.2553 \approx -0.6532 \][/tex]
Now divide by [tex]\(-3\)[/tex]:
[tex]\[ x = \frac{-0.6532}{-3} \approx 0.2177 \][/tex]
Therefore, the exact value of [tex]\(x\)[/tex] is:
[tex]\[ x = \frac{\log_{10}(4) - \log_{10}(18)}{-3} \][/tex]
And the approximate value of [tex]\(x\)[/tex], rounded to the nearest thousandth, is:
[tex]\[ x \approx 0.218 \][/tex]
