Standard +0.3 This requires taking logarithms of both sides after algebraic manipulation (factoring out 5^x), then solving a linear equation in 5^x. It's slightly above average difficulty due to the non-standard form requiring algebraic insight before applying logarithms, but remains a straightforward P3-level question with clear methodology once the manipulation is recognized.
Use laws of indices correctly and solve for \(5^x\) or for \(5^{-x}\) or for \(5^{x-1}\)
M1
Obtain \(5^x\) or for \(5^{-x}\) or for \(5^{x-1}\) in any correct form, e.g. \(5^x = \frac{1}{1-\frac{1}{5}}\)
A1
Use correct method for solving \(5^x = a\), or \(5^{-x} = a\), or \(5^{x-1} = a\), where \(a > 0\)
M1
Obtain answer \(x = 1.14\)
A1
OR
Use an appropriate iterative formula, e.g. \(x_{n+1} = \frac{\ln(5^{x_n}+5)}{\ln 5}\), correctly, at least once
M1
Obtain answer 1.14
A1
Show sufficient iterations to at least 3 d.p. to justify 1.14 to 2 d.p., or show there is a sign change in the interval (1.135, 1.145)
A1
Show there is no other root
A1
[4]
**EITHER** | |
Use laws of indices correctly and solve for $5^x$ or for $5^{-x}$ or for $5^{x-1}$ | M1 |
Obtain $5^x$ or for $5^{-x}$ or for $5^{x-1}$ in any correct form, e.g. $5^x = \frac{1}{1-\frac{1}{5}}$ | A1 |
Use correct method for solving $5^x = a$, or $5^{-x} = a$, or $5^{x-1} = a$, where $a > 0$ | M1 |
Obtain answer $x = 1.14$ | A1 |
**OR** | |
Use an appropriate iterative formula, e.g. $x_{n+1} = \frac{\ln(5^{x_n}+5)}{\ln 5}$, correctly, at least once | M1 |
Obtain answer 1.14 | A1 |
Show sufficient iterations to at least 3 d.p. to justify 1.14 to 2 d.p., or show there is a sign change in the interval (1.135, 1.145) | A1 |
Show there is no other root | A1 | [4]