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Which exponential equation is equivalent to this logarithmic equation?
[tex]\[ \log_2 x = 24 \][/tex]

A. [tex]\(x^2 = 24\)[/tex]

B. [tex]\(2^{24} = x\)[/tex]

C. [tex]\(2^x = 24\)[/tex]

D. [tex]\(x^{24} = 2\)[/tex]

Sagot :

GinnyAnswer
To solve the problem of converting the logarithmic equation [tex]\(\log_2 x = 24\)[/tex] into an equivalent exponential equation, we need to use the properties of logarithms and exponentials.

The logarithmic equation [tex]\(\log_b a = c\)[/tex] can be rewritten as [tex]\(b^c = a\)[/tex]. Here, [tex]\(b\)[/tex] is the base of the logarithm, [tex]\(a\)[/tex] is the argument (or the value inside the logarithm), and [tex]\(c\)[/tex] is the result of the logarithm.

Applying this property to the given equation:

- The base [tex]\(b\)[/tex] is 2.
- The result [tex]\(c\)[/tex] is 24.
- The argument [tex]\(a\)[/tex] is [tex]\(x\)[/tex].

So, using the property of logarithms:

[tex]\[ b^c = a \implies 2^{24} = x \][/tex]

This is the equivalent exponential equation. Therefore, the correct answer is:

B. [tex]\(2^{24} = x\)[/tex]

As we calculated, the value of [tex]\(2^{24}\)[/tex] is 16777216, which confirms our understanding and transformation from the logarithmic form to the exponential form.
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