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Which logarithmic equation is equivalent to this exponential equation?

[tex]\[ 2^x = 32 \][/tex]

A. [tex]\(\log_{32} 2 = x\)[/tex]

B. [tex]\(\log_2 x = 32\)[/tex]

C. [tex]\(\log_{32} x = 2\)[/tex]

D. [tex]\(\log_2 32 = x\)[/tex]

Sagot :

GinnyAnswer
To determine which logarithmic equation is equivalent to the given exponential equation [tex]\(2^x = 32\)[/tex], let's review the relationship between exponential and logarithmic forms.

An exponential equation of the form [tex]\(a^b = c\)[/tex] can be rewritten in logarithmic form as [tex]\(\log_a(c) = b\)[/tex].

Given the exponential equation:
[tex]\[ 2^x = 32 \][/tex]

We need to express this in logarithmic form. According to the logarithmic definition:

1. The base [tex]\(a\)[/tex] of the exponential (which is 2 in this case) becomes the base of the logarithm.
2. The exponent [tex]\(b\)[/tex] (which is [tex]\(x\)[/tex] in this case) becomes the result of the logarithm.
3. The result [tex]\(c\)[/tex] (which is 32 in this case) becomes the argument of the logarithm.

So, applying these rules, we convert [tex]\(2^x = 32\)[/tex] into logarithmic form as:
[tex]\[ \log_2(32) = x \][/tex]

Therefore, the correct logarithmic equation is:
[tex]\[ \log_2 32 = x \][/tex]

Thus, the answer is:
[tex]\[ \log_2 32 = x \][/tex]
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