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QuestionWhich expression equals log⁡BADC\log \frac{B^{A}}{D^{C}}? F. AClog⁡(B−D)A C \log (B-D) G. Alog⁡B−Clog⁡DA \log B-C \log D H. Blog⁡A+Dlog⁡CB \log A+D \log C J. Blog⁡A−Dlog⁡CB \log A-D \log C K. Alog⁡B÷Clog⁡DA \log B \div C \log D

Studdy Solution

STEP 1

Assumptions1. The logarithm is in base10. . The variables A, B, C, and D represent real numbers.

STEP 2

We can use the properties of logarithms to simplify the given expression. The first property we will use is the quotient rule, which states that the logarithm of a quotient is equal to the difference of the logarithms of the numerator and the denominator.
log⁡BAC=log⁡BA−log⁡DC\log \frac{B^{A}}{^{C}} = \log B^{A} - \log D^{C}

STEP 3

Next, we use the power rule of logarithms, which states that the logarithm of a number raised to an exponent is equal to the exponent times the logarithm of the number.
log⁡BA−log⁡DC=Alog⁡B−Clog⁡D\log B^{A} - \log D^{C} = A \log B - C \log D

STEP 4

Now, we compare this expression with the given options. The expression Alog⁡B−Clog⁡DA \log B - C \log D matches with option G.
So, log⁡BAC\log \frac{B^{A}}{^{C}} is equal to Alog⁡B−Clog⁡DA \log B - C \log D.

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