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Answer :
To rewrite the equation [tex]\(2^x = 128\)[/tex] as a logarithmic equation, we need to understand the relationship between exponential and logarithmic forms.
Here's how it works:
1. In an exponential equation like [tex]\(a^b = c\)[/tex], [tex]\(a\)[/tex] is the base, [tex]\(b\)[/tex] is the exponent, and [tex]\(c\)[/tex] is the result.
2. This can be rewritten in logarithmic form as [tex]\(\log_a c = b\)[/tex], where:
- [tex]\(a\)[/tex] is the base of the logarithm,
- [tex]\(c\)[/tex] is the number we're taking the logarithm of,
- [tex]\(b\)[/tex] is the result of the logarithm.
Applying this to the given equation [tex]\(2^x = 128\)[/tex]:
- The base ([tex]\(a\)[/tex]) is 2,
- The result ([tex]\(c\)[/tex]) is 128,
- The exponent ([tex]\(b\)[/tex]), which is what we solve for, is [tex]\(x\)[/tex].
Thus, the logarithmic form of [tex]\(2^x = 128\)[/tex] is [tex]\(\log_2 128 = x\)[/tex].
This correctly identifies that [tex]\(x\)[/tex] is the logarithm to base 2 of 128. Therefore, the correct logarithmic equation is [tex]\(\log_2 128 = x\)[/tex].
Here's how it works:
1. In an exponential equation like [tex]\(a^b = c\)[/tex], [tex]\(a\)[/tex] is the base, [tex]\(b\)[/tex] is the exponent, and [tex]\(c\)[/tex] is the result.
2. This can be rewritten in logarithmic form as [tex]\(\log_a c = b\)[/tex], where:
- [tex]\(a\)[/tex] is the base of the logarithm,
- [tex]\(c\)[/tex] is the number we're taking the logarithm of,
- [tex]\(b\)[/tex] is the result of the logarithm.
Applying this to the given equation [tex]\(2^x = 128\)[/tex]:
- The base ([tex]\(a\)[/tex]) is 2,
- The result ([tex]\(c\)[/tex]) is 128,
- The exponent ([tex]\(b\)[/tex]), which is what we solve for, is [tex]\(x\)[/tex].
Thus, the logarithmic form of [tex]\(2^x = 128\)[/tex] is [tex]\(\log_2 128 = x\)[/tex].
This correctly identifies that [tex]\(x\)[/tex] is the logarithm to base 2 of 128. Therefore, the correct logarithmic equation is [tex]\(\log_2 128 = x\)[/tex].
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