High School

Solve the equation by rewriting in logarithmic form.

15. [tex]2^x = 16[/tex]
[tex]\log_2 16 = x[/tex]
[tex]x = 4[/tex]

16. [tex]4^x - 2 = 7[/tex]

17. [tex]e^x = 3[/tex]

Answer :

Sure! Let's go through each equation step-by-step to find the solutions:

15. Equation: [tex]\(2^x = 16\)[/tex]

To solve this, we rewrite the equation in logarithmic form:
- We know that [tex]\(16 = 2^4\)[/tex].
- This implies [tex]\(2^x = 2^4\)[/tex].

Since the bases are the same, the exponents must be equal:
- Therefore, [tex]\(x = 4\)[/tex].

16. Equation: [tex]\(4^x - 2 = 7\)[/tex]

First, let's move 2 to the other side of the equation:
- [tex]\(4^x = 7 + 2\)[/tex]
- [tex]\(4^x = 9\)[/tex]

Now, we'll rewrite this exponential equation in logarithmic form:
- [tex]\(\log_4 9 = x\)[/tex]

This can be solved using logarithms:
- [tex]\(x = \frac{\log(9)}{\log(4)}\)[/tex]

After evaluating, we find:
- [tex]\(x \approx 1.585\)[/tex]

17. Equation: [tex]\(e^x = 3\)[/tex]

Here, we need to isolate [tex]\(x\)[/tex] using the natural logarithm:
- Taking the natural logarithm (ln) on both sides, we get [tex]\(\ln(e^x) = \ln(3)\)[/tex].

Because [tex]\(\ln(e^x) = x\)[/tex] (due to the properties of logarithms):
- [tex]\(x = \ln(3)\)[/tex]

Evaluating this expression gives us:
- [tex]\(x \approx 1.0986\)[/tex]

These are the solutions:

- For [tex]\(2^x = 16\)[/tex], [tex]\(x = 4\)[/tex].
- For [tex]\(4^x - 2 = 7\)[/tex], [tex]\(x \approx 1.585\)[/tex].
- For [tex]\(e^x = 3\)[/tex], [tex]\(x \approx 1.0986\)[/tex].