Showing all necessary working, solve the equation \(9^x = 3^x + 12\). Give your answer correct to 2 decimal places.
Solution
First, express \(9^x\) in terms of base 3: \(9^x = (3^2)^x = (3^x)^2\).
Let \(u = 3^x\). Then the equation becomes \(u^2 = u + 12\).
Rearrange to form a quadratic equation: \(u^2 - u - 12 = 0\).
Factor the quadratic: \((u - 4)(u + 3) = 0\).
Thus, \(u = 4\) or \(u = -3\). Since \(u = 3^x\) and \(3^x > 0\), we have \(u = 4\).
Therefore, \(3^x = 4\).
Take the natural logarithm of both sides: \(x \ln 3 = \ln 4\).
Solve for \(x\): \(x = \frac{\ln 4}{\ln 3} \approx 1.26186\).
Round to two decimal places: \(x = 1.26\).
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