Moderate -0.3 This is a straightforward exponential equation requiring recognition that 9^x = (3^2)^x = (3^x)^2, leading to a simple quadratic substitution (let y = 3^x). The solution involves basic algebraic manipulation and logarithms, making it slightly easier than average but still requiring proper technique.
State or imply \(u^2 - u - 12(=0)\), or equivalent in \(3^x\)
B1
Need to be convinced they know \(3^{2x}=(3^x)^2\)
Solve for \(u\), or for \(3^x\), and obtain root 4
B1
Use a correct method to solve an equation of the form \(3^x = a\) where \(a > 0\)
M1
Need to see evidence of method. Do not penalise an attempt to use the negative root as well. e.g. \(x\ln 3 = \ln a\), \(x = \log_3 a\). If seen, accept solution of straight forward cases such as \(3^x = 3\), \(x=1\) without working
Obtain final answer \(x = 1.26\) only
A1
The Q asks for 2 dp
4
## Question 2:
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $u^2 - u - 12(=0)$, or equivalent in $3^x$ | B1 | Need to be convinced they know $3^{2x}=(3^x)^2$ |
| Solve for $u$, or for $3^x$, and obtain root 4 | B1 | |
| Use a correct method to solve an equation of the form $3^x = a$ where $a > 0$ | M1 | Need to see evidence of method. Do not penalise an attempt to use the negative root as well. e.g. $x\ln 3 = \ln a$, $x = \log_3 a$. If seen, accept solution of straight forward cases such as $3^x = 3$, $x=1$ without working |
| Obtain final answer $x = 1.26$ **only** | A1 | The Q asks for 2 dp |
| | **4** | |
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2 Showing all necessary working, solve the equation $9 ^ { x } = 3 ^ { x } + 12$. Give your answer correct to 2 decimal places.\\
\hfill \mbox{\textit{CAIE P3 2019 Q2 [4]}}