Covers the primary methods for solving equations where the variable is in the exponent. This includes finding a common base and the more general method of taking the logarithm of both sides.
What is the first method to try when solving an exponential equation?
Try to rewrite both sides of the equation with a common base. If \(b^M = b^N\), then \(M=N\).
Solve for \(x\): \( 5^{2x-1} = 125 \)
Rewrite \(125\) as \(5^3\). Then \(5^{2x-1} = 5^3\), so \(2x-1 = 3\). Solving gives \(2x=4\), so \(x=2\).
Solve for \(x\): \( 4^{x} = 8 \)
Rewrite both sides with base \(2\): \((2^2)^x = 2^3\), which is \(2^{2x} = 2^3\). Therefore, \(2x=3\) and \(x = 3/2\).
What is the general strategy for solving an exponential equation when the bases cannot be made the same?
Isolate the exponential expression and then take the logarithm (common log or natural log) of both sides. Then use the power rule for logs to bring the exponent down and solve for the variable.
Outline the steps to solve \(5^x = 14\).
1. Take the log of both sides: \(\log(5^x) = \log(14)\).
2. Use the power rule: \(x \cdot \log(5) = \log(14)\).
3. Solve for \(x\): \(x = \frac{\log(14)}{\log(5)}\).
Solve for \(x\) in the equation \(e^{x+2} = 9\).
Take the natural logarithm of both sides: \(\ln(e^{x+2}) = \ln(9)\). This simplifies to \(x+2 = \ln(9)\), so \(x = \ln(9) - 2\).
Outline the steps to solve \(3 \cdot 4^{2x} - 5 = 10\).
1. Isolate the exponential part: Add \(5\) to both sides to get \(3 \cdot 4^{2x} = 15\).
2. Divide by \(3\) to get \(4^{2x} = 5\).
3. Take the log of both sides: \(\log(4^{2x}) = \log(5)\).
4. Use the power rule: \(2x \cdot \log(4) = \log(5)\).
5. Solve for \(x\): \(x = \frac{\log(5)}{2\log(4)}\).
How can you solve an equation with a quadratic form, like \(e^{2x} - 6e^x + 8 = 0\)?
Use a substitution. Let \(u = e^x\). The equation becomes \(u^2 - 6u + 8 = 0\). Factor it to \((u-4)(u-2)=0\), so \(u=4\) or \(u=2\). Substitute back: \(e^x=4\) gives \(x=\ln(4)\), and \(e^x=2\) gives \(x=\ln(2)\).
Solve for \(x\): \(2^{x+1} = 5^{1-2x}\).
Take the natural log of both sides: \(\ln(2^{x+1}) = \ln(5^{1-2x})\). Use the power rule: \((x+1)\ln(2) = (1-2x)\ln(5)\). Distribute, group x-terms, and solve for \(x\) to get \(x = \frac{\ln(5)-\ln(2)}{\ln(2)+2\ln(5)}\).
