Question

score on last try: 0 of 1 pts. see details for more. at least one scored part is incorrect. jump to first changable incorrect part. next question get a similar question you can retry this question fill in each box below with an integer or a reduced fraction. (a) \\(\log_{2} 16 = 4\\) can be written in the form \\(2^{a} = b\\) where \\(a = \square\\) and \\(b = \square\\) (b) \\(\log_{5} 25 = 2\\) can be written in the form \\(5^{c} = d\\) where \\(c = \square\\) and \\(d = \square\\) question help: video

Explanation:

Step1: Recall the logarithmic to exponential conversion rule

The logarithmic equation \(\log_{b}x = y\) is equivalent to the exponential equation \(b^{y}=x\).

Step2: Solve part (a)

For \(\log_{2}16 = 4\), using the conversion rule \(b = 2\), \(y=4\), \(x = 16\). So in the form \(2^{A}=B\), we have \(A = 4\) and \(B=16\) (since \(2^{4}=16\)).

Step3: Solve part (b)

For \(\log_{5}25=2\), using the conversion rule \(b = 5\), \(y = 2\), \(x=25\). So in the form \(5^{C}=D\), we have \(C = 2\) and \(D = 25\) (since \(5^{2}=25\)).

Answer:

(a) \(A=\boldsymbol{4}\), \(B=\boldsymbol{16}\)
(b) \(C=\boldsymbol{2}\), \(D=\boldsymbol{25}\)