Standard +0.3 This is a straightforward modulus equation requiring case analysis (3^x ≥ 1 or 3^x < 1) followed by routine algebraic manipulation and logarithms. The exponential substitution y = 3^x simplifies it to linear equations in each case. Slightly above average due to the modulus-exponential combination, but still a standard textbook exercise with clear methodology.
EITHER: State or imply non-modular equation \(2^{\left(3^{x}-1\right)^{2}}=\left(3^{x}\right)^{2}\), or pair of equations \(2\left(3^{x}-1\right)=\pm 3^{x}\)
M1
Obtain \(3^{x}=2\) and \(3^{x}=\frac{2}{3}\) (or \(3^{x+1}=2\))
A1
OR: Obtain \(3^{x}=2\) by solving an equation or by inspection
B1
Obtain \(3^{x}=\frac{2}{3}\) (or \(3^{x+1}=2\)) by solving an equation or by inspection
B1
Use correct method for solving an equation of the form \(3^{x}=a\) (or \(3^{x+1}=a\)), where \(a>0\)
M1
Obtain final answers 0.631 and −0.369
A1
[4]
**EITHER:** State or imply non-modular equation $2^{\left(3^{x}-1\right)^{2}}=\left(3^{x}\right)^{2}$, or pair of equations $2\left(3^{x}-1\right)=\pm 3^{x}$ | M1 |
Obtain $3^{x}=2$ and $3^{x}=\frac{2}{3}$ (or $3^{x+1}=2$) | A1 |
**OR:** Obtain $3^{x}=2$ by solving an equation or by inspection | B1 |
Obtain $3^{x}=\frac{2}{3}$ (or $3^{x+1}=2$) by solving an equation or by inspection | B1 |
Use correct method for solving an equation of the form $3^{x}=a$ (or $3^{x+1}=a$), where $a>0$ | M1 |
Obtain final answers 0.631 and −0.369 | A1 | [4]