Example 1: Solve \(e^{2x} = 6\)
The expression \(e^{2x} = 6\) can be written as a natural logarithm as the base is e, the exponent is 2x, and the answer to the exponential is 6.
So as a natural logarithm, it could be written as ln (6) = 2x.
Therefore, \(\frac{\ln(6)}{2} = 0.896 (3 s.f)\)
Example 2: Solve \(e^{x+3} = 10\)
The expression \(e^{x+3}\) can be written as a logarithm, whereby the base is e; the exponent is x + 3, and the answer to the exponential is 10.
\(\ln(10) = x + 3\)
Therefore, \(x = \ln(10) - 3 = -0.697(3 s.f)\)
Example 3: Solve \(e^{\ln(x^3)} = 8\)
As the exponential and logarithms are inverse functions, the e and Ln will cancel each other.
Therefore, \(x^3 = 8; x = 2\)
Example 4: Solve \(\ln(x+1) = 1.4\)
To get x on its own, we need to convert the logarithm to an exponential where the base is e, the exponent is 1.4, and the answer to the exponential is x + 1.
Therefore, \(e^{1.4} = x+1\) and \(x = e^{1.4} -1 = 3.06(3 s.f)\)
Example 5: Solve \(2\ln(6) + \ln(2) - \ln(4) = x\)
1. Due to the power logarithm rule, \(2\ln(6)\) can be written as \(\ln(6^2) = \ln(36)\)
Therefore, \(\ln(36) +\ln(2) - \ln(4) = x\)
2. Using the product and quotient rule, we can do this further:
\(\ln(36 \cdot 2) - \ln(4) = x\)
\(\ln(\frac{36 \cdot 2}{4}) = x\)
\(\ln(\frac{72}{4}) = \ln(18) = x = 2.89 (3 s.f)\)