Question

convert the exponential equations into logarithmic form: (a) $2^4 = 16$ is equivalent to $log_2 c = d$. $c = square$ and $d = square$ (b) $125 = 5^3$ is equivalent to $log_5 e = f$. $e = square$ and $f = square$ (c) $10^2 = 100$ is equivalent to $log_{10} g = h$. $g = square$ and $h = square$ (d) $0.1 = 10^{-1}$ is equivalent to $log_{10} j = k$. $j = square$ and $k = square$ question help: video

Explanation:

Step1: Recall the relationship between exponential and logarithmic forms

The general relationship is \(a^b = c\) is equivalent to \(\log_a c = b\).

Step2: Solve part (a)

For \(2^4 = 16\), comparing with \(\log_2 C = D\), we have \(a = 2\), \(b = 4\), \(c = 16\). So \(C = 16\) and \(D = 4\).

Step3: Solve part (b)

For \(125 = 5^3\), which can be written as \(5^3 = 125\). Comparing with \(\log_5 E = F\), we have \(a = 5\), \(b = 3\), \(c = 125\). So \(E = 125\) and \(F = 3\).

Step4: Solve part (c)

For \(10^2 = 100\), comparing with \(\log_{10} G = H\), we have \(a = 10\), \(b = 2\), \(c = 100\). So \(G = 100\) and \(H = 2\).

Step5: Solve part (d)

For \(0.1 = 10^{-1}\), which can be written as \(10^{-1} = 0.1\). Comparing with \(\log_{10} J = K\), we have \(a = 10\), \(b = -1\), \(c = 0.1\). So \(J = 0.1\) and \(K = -1\).

Answer:

(a) \(C = 16\), \(D = 4\)
(b) \(E = 125\), \(F = 3\)
(c) \(G = 100\), \(H = 2\)
(d) \(J = 0.1\), \(K = -1\)