Moderate -0.8 This is a straightforward exponential equation requiring standard logarithm techniques: rewrite bases as powers of 2 and 3, apply log rules, and solve the resulting linear equation. It's a single-step problem testing routine manipulation skills with no conceptual challenges, making it easier than the typical A-level question.
Use law of the logarithm of a product or a quotient or a power
\*M1
Obtain a correct linear equation in any form
A1
e.g. \(\ln 2 + (2x-1)\ln 3 = (x+1)\ln 4\) or \(\log_2 2 + (2x-1)\log_2 3 = (2x+2)\log_2 2\)
Solve for \(x\)
DM1
Allow for unsimplified expression \(x = \ldots\); Allow M1 M1 for \(x = 1.45\) from \(6^{2x-1} = 4^{x+1}\)
Obtain answer \(x = 2.21\)
A1
The question asks for 2 dp
Alternative method for question 1:
Answer
Marks
Guidance
Answer
Marks
Guidance
Correct use of indices to obtain \(2.25^x = 6\) or \(1.5^{2x} = 6\)
M1 A1
Correct use of logarithms to solve for \(x\)
M1
Allow solution of \(2.25^x = 6\) by trial and improvement as far as \(2.2\ldots\)
Obtain answer \(x = 2.21\)
A1
Need to see an intermediate step / sequence of iterations
4
## Question 1:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use law of the logarithm of a product or a quotient or a power | \*M1 | |
| Obtain a correct linear equation in any form | A1 | e.g. $\ln 2 + (2x-1)\ln 3 = (x+1)\ln 4$ or $\log_2 2 + (2x-1)\log_2 3 = (2x+2)\log_2 2$ |
| Solve for $x$ | DM1 | Allow for unsimplified expression $x = \ldots$; Allow M1 M1 for $x = 1.45$ from $6^{2x-1} = 4^{x+1}$ |
| Obtain answer $x = 2.21$ | A1 | The question asks for 2 dp |
**Alternative method for question 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct use of indices to obtain $2.25^x = 6$ or $1.5^{2x} = 6$ | M1 A1 | |
| Correct use of logarithms to solve for $x$ | M1 | Allow solution of $2.25^x = 6$ by trial and improvement as far as $2.2\ldots$ |
| Obtain answer $x = 2.21$ | A1 | Need to see an intermediate step / sequence of iterations |
| | **4** | |