Sure, let's solve the equation step-by-step:
Given the equation:
[tex]\[
\log_3{8} - 3 \log_3{t} = 3
\][/tex]
1. Combine the logarithms using properties of logarithms:
The property we will use is that [tex]\(a \log_b{c} = \log_b{c^a}\)[/tex]. Applying this property:
[tex]\[
\log_3{8} - \log_3{t^3} = 3
\][/tex]
2. Combine the logarithmic terms on the left-hand side:
Another property of logarithms is [tex]\(\log_b{a} - \log_b{c} = \log_b{\frac{a}{c}}\)[/tex]. Applying this property:
[tex]\[
\log_3{\left(\frac{8}{t^3}\right)} = 3
\][/tex]
3. Exponentiate both sides to remove the logarithm:
To remove the logarithm, rewrite the equation in exponential form. If [tex]\(\log_b{a} = c\)[/tex], then [tex]\(a = b^c\)[/tex]. Thus:
[tex]\[
\frac{8}{t^3} = 3^3
\][/tex]
4. Simplify the equation:
Calculate [tex]\(3^3\)[/tex]:
[tex]\[
3^3 = 27
\][/tex]
So the equation becomes:
[tex]\[
\frac{8}{t^3} = 27
\][/tex]
5. Solve for [tex]\(t^3\)[/tex]:
To isolate [tex]\(t^3\)[/tex], we can multiply both sides by [tex]\(t^3\)[/tex] and then divide both sides by 27:
[tex]\[
8 = 27 t^3
\][/tex]
[tex]\[
t^3 = \frac{8}{27}
\][/tex]
6. Take the cube root to solve for [tex]\(t\)[/tex]:
Finally, take the cube root of both sides:
[tex]\[
t = \sqrt[3]{\frac{8}{27}}
\][/tex]
Let's express the result found:
[tex]\[
t^3 = \frac{8}{27}
\][/tex]
[tex]\[
t = \sqrt[3]{\frac{8}{27}} = \left(\frac{8}{27}\right)^{\frac{1}{3}} \approx 0.6667
\][/tex]
Therefore, we found:
[tex]\[
t^3 \approx 0.2963 \quad \text{and} \quad t \approx 0.6667
\][/tex]
So the solution is:
[tex]\[
t \approx 0.6667
\][/tex]
