Answer :
Certainly! Let's solve the equation [tex]\(10^{\frac{2z}{3}} = 15\)[/tex] for [tex]\(z\)[/tex].
### Step-by-Step Solution:
1. Convert to Logarithmic Form:
To solve for [tex]\(z\)[/tex], we first need to express the given equation in logarithmic form. We know that:
[tex]\[ 10^{\frac{2z}{3}} = 15 \][/tex]
Taking the logarithm (base-10) of both sides, we get:
[tex]\[ \log_{10}\left(10^{\frac{2z}{3}}\right) = \log_{10}(15) \][/tex]
2. Apply the Power Rule of Logarithms:
Using the power rule [tex]\(\log_{10}(a^b) = b \log_{10}(a)\)[/tex], we can rewrite the left side as:
[tex]\[ \frac{2z}{3} \cdot \log_{10}(10) = \log_{10}(15) \][/tex]
Since [tex]\(\log_{10}(10) = 1\)[/tex], this simplifies to:
[tex]\[ \frac{2z}{3} = \log_{10}(15) \][/tex]
3. Solve for [tex]\(z\)[/tex]:
Now, we can isolate [tex]\(z\)[/tex] by multiplying both sides of the equation by [tex]\(\frac{3}{2}\)[/tex]:
[tex]\[ z = \left(\frac{3}{2}\right) \log_{10}(15) \][/tex]
Thus, the exact solution for [tex]\(z\)[/tex] expressed as a logarithm in base-10 is:
[tex]\[ z = \left(\frac{3}{2}\right) \log_{10}(15) \][/tex]
### Approximate the Value of [tex]\(z\)[/tex]:
To approximate the value of [tex]\(z\)[/tex], compute [tex]\(\log_{10}(15)\)[/tex] and then multiply by [tex]\(\frac{3}{2}\)[/tex]:
1. Calculate [tex]\(\log_{10}(15)\)[/tex]:
[tex]\[ \log_{10}(15) \approx 1.176 \][/tex]
2. Multiply by [tex]\(\frac{3}{2}\)[/tex]:
[tex]\[ z \approx \left(\frac{3}{2}\right) \times 1.176 = 1.764 \][/tex]
Therefore, rounding to the nearest thousandth, we get:
[tex]\[ z \approx 1.764 \][/tex]
In summary:
[tex]\[ z = \left(\frac{3}{2}\right) \log_{10}(15) \][/tex]
and the approximate value of [tex]\(z\)[/tex] is:
[tex]\[ z \approx 1.764 \][/tex]
### Step-by-Step Solution:
1. Convert to Logarithmic Form:
To solve for [tex]\(z\)[/tex], we first need to express the given equation in logarithmic form. We know that:
[tex]\[ 10^{\frac{2z}{3}} = 15 \][/tex]
Taking the logarithm (base-10) of both sides, we get:
[tex]\[ \log_{10}\left(10^{\frac{2z}{3}}\right) = \log_{10}(15) \][/tex]
2. Apply the Power Rule of Logarithms:
Using the power rule [tex]\(\log_{10}(a^b) = b \log_{10}(a)\)[/tex], we can rewrite the left side as:
[tex]\[ \frac{2z}{3} \cdot \log_{10}(10) = \log_{10}(15) \][/tex]
Since [tex]\(\log_{10}(10) = 1\)[/tex], this simplifies to:
[tex]\[ \frac{2z}{3} = \log_{10}(15) \][/tex]
3. Solve for [tex]\(z\)[/tex]:
Now, we can isolate [tex]\(z\)[/tex] by multiplying both sides of the equation by [tex]\(\frac{3}{2}\)[/tex]:
[tex]\[ z = \left(\frac{3}{2}\right) \log_{10}(15) \][/tex]
Thus, the exact solution for [tex]\(z\)[/tex] expressed as a logarithm in base-10 is:
[tex]\[ z = \left(\frac{3}{2}\right) \log_{10}(15) \][/tex]
### Approximate the Value of [tex]\(z\)[/tex]:
To approximate the value of [tex]\(z\)[/tex], compute [tex]\(\log_{10}(15)\)[/tex] and then multiply by [tex]\(\frac{3}{2}\)[/tex]:
1. Calculate [tex]\(\log_{10}(15)\)[/tex]:
[tex]\[ \log_{10}(15) \approx 1.176 \][/tex]
2. Multiply by [tex]\(\frac{3}{2}\)[/tex]:
[tex]\[ z \approx \left(\frac{3}{2}\right) \times 1.176 = 1.764 \][/tex]
Therefore, rounding to the nearest thousandth, we get:
[tex]\[ z \approx 1.764 \][/tex]
In summary:
[tex]\[ z = \left(\frac{3}{2}\right) \log_{10}(15) \][/tex]
and the approximate value of [tex]\(z\)[/tex] is:
[tex]\[ z \approx 1.764 \][/tex]
