Moderate -0.8 This is a straightforward exponential equation requiring basic manipulation (factoring out 3^x) and simple logarithms. The algebraic trick of rewriting 3^(x+2) as 9ยท3^x makes it a linear equation in 3^x, which is simpler than typical logarithm problems. Below average difficulty as it's more routine than the standard C3 multi-part question.
Use laws of indices correctly and solve a linear equation for \(3^x\), or for \(3^{-x}\)
M1
Obtain \(3^x\), or \(3^{-x}\) in any correct form, e.g. \(3^x = \frac{3^2}{(3^2-1)}\)
A1
Use correct method for solving \(3^x = a\) for \(x\), where \(a > 0\)
M1
Obtain answer \(x = 0.107\)
A1
[4]
OR:
Answer
Marks
State an appropriate iterative formula, e.g. \(x_{n+1} = \frac{\ln(3^x + 9)}{\ln 3} - 2\)
B1
Use the formula correctly at least once
M1
Obtain answer \(x = 0.107\)
A1
Show that the equation has no other root but 0.107
A1
[For the solution 0.107 with no relevant working, award B1 and a further B1 if 0.107 is shown to be the only root.]
**EITHER:**
| Use laws of indices correctly and solve a linear equation for $3^x$, or for $3^{-x}$ | M1 |
| Obtain $3^x$, or $3^{-x}$ in any correct form, e.g. $3^x = \frac{3^2}{(3^2-1)}$ | A1 |
| Use correct method for solving $3^x = a$ for $x$, where $a > 0$ | M1 |
| Obtain answer $x = 0.107$ | A1 |
| [4] |
**OR:**
| State an appropriate iterative formula, e.g. $x_{n+1} = \frac{\ln(3^x + 9)}{\ln 3} - 2$ | B1 |
| Use the formula correctly at least once | M1 |
| Obtain answer $x = 0.107$ | A1 |
| Show that the equation has no other root but 0.107 | A1 |
| [For the solution 0.107 with no relevant working, award B1 and a further B1 if 0.107 is shown to be the only root.] |