Given a very basic exponential function, \({\color{{green}}y}=B^{\color{{red}}x}\), the inverse to this function is called a logarithm and we would write it: \({\color{{red}}x}=\log_B({\color{{green}}y})\). Notice that the input/output switch and that the Base is either the base of the exponential or the base of the logarithm. You will need to be able to transform an exponential function to a logarithmic function and vice versa.
Mini Examples: Rewrite each equation as either the logarithmic version or exponential (whichever is not given).
- \(3=(1.15)^t\)
- \(100=\log_{{10}}(x)\)
- \(2^{{8x}}=y\)
- \(\log_3(2)=x\)
- \(A=P(1.01)^x\)
- \(3-0.2x = \log_5(y-1)\)
Solution
- \(3=(1.15)^t\rightarrow \log_{1.15}(3) = t\)
- \(100=\log_{{10}}(x)\rightarrow 10^{100} = x\)
- \(2^{{8x}}=y\rightarrow 8x=\log_2(y)\)
- \(\log_3(2)=x\rightarrow 2=3^x\)
- This is not strictly written in the exponential form, so we first need to divide by \(P\):\(A=P(1.01)^x\rightarrow \frac{A}{P} = 1.01^x\rightarrow \log_{1.01}\left(\frac{A}{P}\right)=x\)
- \(3-0.2x = \log_5(y-1)\rightarrow 5^{3-0.2x}=y-1\)