Notes
Logarithmic and Exponential Form
\(\log_{b}a=c\,\, \Leftrightarrow \,\, b^c=a\)
Practice Problems
\(\textbf{1)}\) \(\log_{3}9\)
The answer is \(2\)
\(\,\,\,\,\,\log_{3}9=x\)
\(\,\,\,\,\,3^x=9\)
\(\,\,\,\,\,3^x=3^{2}\)
\(\,\,\,\,\,x=2\)
\(\textbf{2)}\) \(\log_{3}3\)
The answer is \(1\)
\(\,\,\,\,\,\log_{3}3=x\)
\(\,\,\,\,\,3^x=3\)
\(\,\,\,\,\,3^x=3^{1}\)
\(\,\,\,\,\,x=1\)
\(\textbf{3)}\) \(\log_{2}8\)
The answer is \(3\)
\(\,\,\,\,\,\log_{2}8=x\)
\(\,\,\,\,\,2^x=8\)
\(\,\,\,\,\,2^x=2^{3}\)
\(\,\,\,\,\,x=3\)
\(\textbf{4)}\) \(\log_{4}16\)
The answer is \(2\)
\(\,\,\,\,\,\log_{4}16=x\)
\(\,\,\,\,\,4^x=16\)
\(\,\,\,\,\,4^x=4^2\)
\(\,\,\,\,\,x=2\)
\(\textbf{5)}\) \(\log_{6}6\)
The answer is \(1\)
\(\,\,\,\,\,\log_{6}6=x\)
\(\,\,\,\,\,6^x=6\)
\(\,\,\,\,\,6^x=6^1\)
\(\,\,\,\,\,x=1\)
\(\textbf{6)}\) \(\log_{2}32\)
The answer is \(5\)
\(\,\,\,\,\,\log_{2}32=x\)
\(\,\,\,\,\,2^x=32\)
\(\,\,\,\,\,2^x=2^5\)
\(\,\,\,\,\,x=5\)
\(\textbf{7)}\) \(\log_{6}1\)
The answer is \(0\)
\(\,\,\,\,\,\log_{6}1=x\)
\(\,\,\,\,\,6^x=1\)
\(\,\,\,\,\,6^x=6^0\)
\(\,\,\,\,\,x=0\)
\(\textbf{8)}\) \(\log_{2}\frac{1}{2}\)
The answer is \(-1\)
\(\,\,\,\,\,\log_{2}\frac{1}{2}=x\)
\(\,\,\,\,\,2^x=\frac{1}{2}\)
\(\,\,\,\,\,2^x=2^{-1}\)
\(\,\,\,\,\,x=-1\)
\(\textbf{9)}\) \(\log_{2}\frac{1}{4}\)
The answer is \(-2\)
\(\,\,\,\,\,\log_{2}\frac{1}{4}=x\)
\(\,\,\,\,\,2^x=\frac{1}{4}\)
\(\,\,\,\,\,2^x=2^{-2}\)
\(\,\,\,\,\,x=-2\)
\(\textbf{10)}\) \(\log_{4}2\)
The answer is \(\frac{1}{2}\)
\(\,\,\,\,\,\log_{4}2=x\)
\(\,\,\,\,\,4^x=2\)
\(\,\,\,\,\,\left(2^2\right)^x=2^1\)
\(\,\,\,\,\,2^{2x}=2^1\)
\(\,\,\,\,\,2x=1\)
\(\,\,\,\,\,x=\frac{1}{2}\)
\(\textbf{11)}\) \(\log_{7}\frac{1}{49}\)
The answer is \(-2\)
\(\,\,\,\,\,\log_{7}\frac{1}{49}=x\)
\(\,\,\,\,\,7^x=\frac{1}{49}\)
\(\,\,\,\,\,7^x=7^{-2}\)
\(\,\,\,\,\,x=-2\)
\(\textbf{12)}\) \(\log_{0.5}4\)
The answer is \(-2\)
\(\,\,\,\,\,\log_{0.5}4=x\)
\(\,\,\,\,\,0.5^x=4\)
\(\,\,\,\,\,2^{-1x}=2^{2}\)
\(\,\,\,\,\,-1x=2\)
\(\,\,\,\,\,x=-2\)
\(\textbf{13)}\) \(\log_{\frac{1}{5}}125\)
The answer is \(-3\)
\(\,\,\,\,\,\log_{\frac{1}{5}}125=x\)
\(\,\,\,\,\,\left(\frac{1}{5}\right)^x=125\)
\(\,\,\,\,\,5^{-1x}=5^{3}\)
\(\,\,\,\,\,-1x=3\)
\(\,\,\,\,\,x=-3\)
\(\textbf{14)}\) \(\log_{5}\frac{1}{5}\)
The answer is \(-1\)
\(\,\,\,\,\,\log_{5}\frac{1}{5}=x\)
\(\,\,\,\,\,5^x=\frac{1}{5}\)
\(\,\,\,\,\,5^x=5^{-1}\)
\(\,\,\,\,\,x=-1\)
\(\textbf{15)}\) \(\log_{36}6\)
The answer is \(\frac{1}{2}\)
\(\,\,\,\,\,\log_{36}6=x\)
\(\,\,\,\,\,36^x=6\)
\(\,\,\,\,\,6^{2x}=6^{1}\)
\(\,\,\,\,\,2x=1\)
\(\,\,\,\,\,x=\frac{1}{2}\)
\(\textbf{16)}\) \(\log_{12}12\)
The answer is \(1\)
\(\,\,\,\,\,\log_{12}12=x\)
\(\,\,\,\,\,12^x=12\)
\(\,\,\,\,\,12^x=12^1\)
\(\,\,\,\,\,x=1\)
\(\textbf{17)}\) \(\log_{4}1\)
The answer is \(0\)
\(\,\,\,\,\,\log_{4}1=x\)
\(\,\,\,\,\,4^x=1\)
\(\,\,\,\,\,4^x=4^0\)
\(\,\,\,\,\,x=0\)
\(\textbf{18)}\) \(\log_{7}1\)
The answer is \(0\)
\(\,\,\,\,\,\log_{7}1=x\)
\(\,\,\,\,\,7^x=1\)
\(\,\,\,\,\,7^x=7^0\)
\(\,\,\,\,\,x=0\)
\(\textbf{19)}\) \(\log_{\frac{1}{4}}2\)
The answer is \(-\frac{1}{2}\)
\(\,\,\,\,\,\log_{\frac{1}{4}}2=x\)
\(\,\,\,\,\,\left(\frac{1}{4}\right)^x=2\)
\(\,\,\,\,\,2^{-2x}=2^1\)
\(\,\,\,\,\,-2x=1\)
\(\,\,\,\,\,x=-\frac{1}{2}\)
\(\textbf{20)}\) \(\log_{8}4\)
The answer is \(\frac{2}{3}\)
\(\,\,\,\,\,\log_{8}4=x\)
\(\,\,\,\,\,8^x=4\)
\(\,\,\,\,\,2^{3x}=2^2\)
\(\,\,\,\,\,3x=2\)
\(\,\,\,\,\,x=\frac{2}{3}\)
\(\textbf{21)}\) \(\log_{4}\frac{1}{64}\)
The answer is \(-3\)
\(\,\,\,\,\,\log_{4}\frac{1}{64}=x\)
\(\,\,\,\,\,4^x=\frac{1}{64}\)
\(\,\,\,\,\,2^{2x}=2^{-6}\)
\(\,\,\,\,\,2x=-6\)
\(\,\,\,\,\,x=-3\)
\(\textbf{22)}\) \(\log_{125}5\)
The answer is \(\frac{1}{3}\)
\(\,\,\,\,\,\log_{125}5=x\)
\(\,\,\,\,\,125^x=5\)
\(\,\,\,\,\,5^{3x}=5^1\)
\(\,\,\,\,\,3x=1\)
\(\,\,\,\,\,x=\frac{1}{3}\)
\(\textbf{23)}\) \(\log_{\frac{1}{27}}3\)
The answer is \(-\frac{1}{3}\)
\(\,\,\,\,\,\log_{\frac{1}{27}}3=x\)
\(\,\,\,\,\,\left(\frac{1}{27}\right)^x=3\)
\(\,\,\,\,\,3^{-3x}=3^1\)
\(\,\,\,\,\,-3x=1\)
\(\,\,\,\,\,x=-\frac{1}{3}\)
\(\textbf{24)}\) \(\log 1\)
The answer is \(0\)
\(\,\,\,\,\,\log 1=x\)
\(\,\,\,\,\,10^x=1\)
\(\,\,\,\,\,10^x=10^0\)
\(\,\,\,\,\,x=0\)
\(\textbf{25)}\) \(\log_{5} \frac{1}{\sqrt{5}}\)
The answer is \(-\frac{1}{2}\)
\(\,\,\,\,\,\log_{5} \frac{1}{\sqrt{5}}=x\)
\(\,\,\,\,\,5^x=\frac{1}{\sqrt{5}}\)
\(\,\,\,\,\,5^x=5^{-1/2}\)
\(\,\,\,\,\,x=-\frac{1}{2}\)
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