Standard +0.8 This requires handling absolute value cases systematically, solving two exponential equations (one yielding x=0 trivially, the other requiring logarithms), and checking validity of solutions. The absolute value adds conceptual complexity beyond standard exponential equations, but the algebraic manipulation remains straightforward once cases are identified.
State or imply non-modular equation \(4^2(5^x-1)^2=(5^x)^2\) or pair of equations \(4(5^x-1)=\pm5^x\)
M1
Obtain \(5^x=\frac{4}{3}\) and \(5^x=\frac{4}{5}\) (or \(5^{x+1}=4\))
A1
Use correct method for solving an equation of the form \(5^x=a\), or \(5^{x+1}=b\) where \(a>0\), or \(b>0\)
M1
Obtain answers \(x=0.179\) and \(x=-0.139\)
A1
Alternative method for Question 1:
Answer
Marks
Guidance
Answer
Mark
Guidance
Obtain \(5^x=\frac{4}{3}\) by solving an equation
B1
Obtain \(5^x=\frac{4}{5}\) (or \(5^{x+1}=4\)) by solving an equation
B1
Use correct method for solving an equation of the form \(5^x=a\), or \(5^{x+1}=b\) where \(a>0\), or \(b>0\)
M1
Obtain answers \(x=0.179\) and \(x=-0.139\)
A1
Total: 4 marks
## Question 1:
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply non-modular equation $4^2(5^x-1)^2=(5^x)^2$ or pair of equations $4(5^x-1)=\pm5^x$ | M1 | |
| Obtain $5^x=\frac{4}{3}$ and $5^x=\frac{4}{5}$ (or $5^{x+1}=4$) | A1 | |
| Use correct method for solving an equation of the form $5^x=a$, or $5^{x+1}=b$ where $a>0$, or $b>0$ | M1 | |
| Obtain answers $x=0.179$ and $x=-0.139$ | A1 | |
**Alternative method for Question 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain $5^x=\frac{4}{3}$ by solving an equation | B1 | |
| Obtain $5^x=\frac{4}{5}$ (or $5^{x+1}=4$) by solving an equation | B1 | |
| Use correct method for solving an equation of the form $5^x=a$, or $5^{x+1}=b$ where $a>0$, or $b>0$ | M1 | |
| Obtain answers $x=0.179$ and $x=-0.139$ | A1 | |
**Total: 4 marks**