Solve $2^{2x}-9\times2^{x}+8=0$.
- Substitutes $u=2^x$ to obtain $u^2-9u+8=0$M1
- $(u-1)(u-8)=0$ so $u=1$ or $u=8$A1
- Solves $2^x=1$ and $2^x=8$M1
- $x=0$ or $x=3$A1
- Answer: $x=0$ or $x=3$
- Examiner tip: Do not stop at $u=1,8$: the question asks for $x$.
Solve $\log_2(x+1)+\log_2(x-1)=3$.
- Combines logs: $\log_2\big((x+1)(x-1)\big)=3$M1
- Converts: $x^2-1=2^3$M1
- $x^2=9$ so $x=\pm3$ (or $x=3$ directly, having stated $x>1$)A1
- Rejects $x=-3$ since $\log_2(x-1)$ (and $\log_2(x+1)$) is undefined for $x=-3$ (or states the domain $x>1$)R1
- $x=3$A1
- Answer: $x=3$
- Examiner tip: Always check solutions against the domain of the original logarithms.
Let $x>0$.
(a) Show that $\log_9 x=\tfrac12\log_3 x$. [2]
(b) Hence, or otherwise, solve $\log_3 x+\log_9 x+\log_{27}x=11$. [4]
- Uses change of base: $\log_9x=\dfrac{\log_3x}{\log_39}$M1
- $\log_39=2$, so $\log_9x=\tfrac12\log_3x$ AGA1
- Writes $\log_{27}x=\tfrac13\log_3x$ (similar method)M1
- $\left(1+\tfrac12+\tfrac13\right)\log_3x=\tfrac{11}{6}\log_3x=11$A1
- $\log_3x=6$A1
- $x=3^6=729$A1
- Answer: $x=729$
- Examiner tip: In a 'show that' question every step must be visible: state $\log_3 9=2$ explicitly.
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(a) Simplify $\dfrac{(3a^{2}b^{-1})^{3}}{9a^{4}b^{-5}}$, giving your answer in the form $ka^{m}b^{n}$, where $k,m,n\in\mathbb{Z}$. [2]
(b) Find the exact value of $\left(\dfrac{16}{81}\right)^{-\frac{3}{4}}$. [2]
- (a) Correctly cubes the bracket: $27a^{6}b^{-3}$ seen (or equivalent such as $\frac{27a^6}{b^3}$)M1
- $3a^{2}b^{2}$ (all three of $k=3$, $m=2$, $n=2$ correct)A1
- (b) Deals with the negative power by inverting OR with the fourth root: $\left(\frac{81}{16}\right)^{\frac34}$ or $\left(\frac{2}{3}\right)^{-3}$ or $\frac{\sqrt[4]{81}}{\sqrt[4]{16}}=\frac32$ seenM1
- $\dfrac{27}{8}$ (accept $3\tfrac{3}{8}$ or $3.375$)A1
- Answer: (a) $3a^{2}b^{2}$ (b) $\dfrac{27}{8}$
- Examiner tip: In (a) the $-1$ power must be cubed too: $(b^{-1})^3=b^{-3}$, and dividing by $b^{-5}$ adds 5 to the power, so $b^{2}$, not $b^{-8}$.
The pH of a solution is given by $\mathrm{pH}=-\log_{10}C$, where $C$ is the hydrogen ion concentration in moles per litre.
(a) Find the pH of a solution with $C=3.2\times10^{-5}$. [2]
(b) A solution has a pH of $2.7$. Find $C$, giving your answer in the form $a\times10^{k}$ where $1\le a<10$ and $k\in\mathbb{Z}$. [2]
(c) Solution A has pH $3.1$ and solution B has pH $5.6$. Find how many times greater the hydrogen ion concentration of A is than that of B. [2]
- (a) Substitutes: $-\log_{10}(3.2\times10^{-5})$M1
- $4.49$ (accept $4.49485\ldots$; accept $4.5$ only if $4.49\ldots$ seen first)A1
- (b) $C=10^{-2.7}$ seen (or $\log_{10}C=-2.7$)M1
- $2.00\times10^{-3}$ (accept $1.995\times10^{-3}$, $2\times10^{-3}$)A1
- (c) $\dfrac{10^{-3.1}}{10^{-5.6}}$ or $10^{5.6-3.1}$ or $10^{2.5}$ seenM1
- $316$ (accept $316.2\ldots$; award M1A0 for $\frac{1}{316}$ or $0.00316$)A1
- Answer: (a) $4.49$ (b) $2.00\times10^{-3}$ (c) $316$ times
- Examiner tip: In (c) the difference in pH is a power of ten, not a multiplier: the ratio is $10^{2.5}\approx316$, not $2.5$ or $5.6\div3.1$.
Let $\log_a 2=p$ and $\log_a 5=q$, where $a>0$, $a\neq1$.
Write each of the following in terms of $p$ and $q$.
(a) $\log_a 40$ [2]
(b) $\log_a 0.08$ [2]
(c) $\log_{a^{2}} 50$ [2]
- (a) Writes $40$ as a product of powers of 2 and 5 and applies a log law: $\log_a 8+\log_a 5$ or $3\log_a2+\log_a5$M1
- $3p+q$A1
- (b) $0.08=\frac{2}{25}$ (or $\frac{8}{100}$) used with the quotient law: $\log_a2-2\log_a5$ or $3\log_a2-2(\log_a2+\log_a5)$M1
- $p-2q$A1
- (c) Uses change of base: $\dfrac{\log_a 50}{\log_a a^{2}}$ (or $\frac{1}{2}\log_a 50$)M1
- $\dfrac{p+2q}{2}$ (accept $\frac12p+q$)A1
- Answer: (a) $3p+q$ (b) $p-2q$ (c) $\dfrac{p+2q}{2}$
- Examiner tip: In (c) $\log_{a^2}50$ is not $2\log_a50$: changing to base $a$ divides by $\log_a a^{2}=2$.
Solve the equation $3^{2x-1}=5^{x+2}$, giving your answer in the form $x=\dfrac{\ln m}{\ln n}$, where $m,n\in\mathbb{Q}$.
- Takes logs of both sides (any base): $\ln 3^{2x-1}=\ln5^{x+2}$M1
- $(2x-1)\ln3=(x+2)\ln5$A1
- Expands and collects the $x$ terms: $x(2\ln3-\ln5)=2\ln5+\ln3$ (or equivalent)M1
- Uses log laws to combine: $2\ln3-\ln5=\ln\frac{9}{5}$ or $2\ln5+\ln3=\ln75$M1
- $x=\dfrac{\ln75}{\ln\frac{9}{5}}$ (accept $\dfrac{\ln75}{\ln1.8}$; do not accept a decimal only, $x\approx7.35$, for this mark)A1
- Answer: $x=\dfrac{\ln 75}{\ln \frac{9}{5}}\ (\approx7.35)$
- Examiner tip: Brackets matter: $\ln3^{2x-1}=(2x-1)\ln3$, and writing $2x-1\ln3$ usually leads to a lost A1 and a wrong collection of terms.
The population, $P$, of insects in a laboratory colony $t$ days after observation began is modelled by $P=Ae^{kt}$, where $A$ and $k$ are constants.
The graph shows $\ln P$ plotted against $t$. It is a straight line passing through the points $(0,\,1.2)$ and $(5,\,3.2)$.
(a) Show that $\ln P=\ln A+kt$. [1]
(b) Find the value of $k$. [1]
(c) Find the value of $A$. [2]
(d) Find the time taken for the population to reach 50 insects. [2]
- (a) Takes ln of both sides and uses the product law: $\ln P=\ln A+\ln e^{kt}=\ln A+kt$ (must see the $\ln A+\ln e^{kt}$ step; AG)A1
- (b) $k=\dfrac{3.2-1.2}{5-0}=0.4$A1
- (c) $\ln A=1.2$ (the vertical intercept)M1
- $A=e^{1.2}=3.32$ (accept $3.3201\ldots$)A1
- (d) $50=e^{1.2}e^{0.4t}$ or $\ln50=1.2+0.4t$ seenM1
- $t=6.78$ days (accept $6.7800\ldots$; award A1 for $6.77$–$6.79$ from a rounded $A$)A1
- Answer: (a) shown (b) $k=0.4$ (c) $A=3.32$ (d) $6.78$ days
- Examiner tip: The intercept of the line is $\ln A$, not $A$: giving $A=1.2$ loses the A1 in (c) and carries through to a wrong time in (d).
Solve the equation $\log_2 x-3\log_x 2=2$, where $x>0$, $x\neq1$.
- Uses change of base to write $\log_x2=\dfrac{1}{\log_2x}$ (or writes both terms in a common base, e.g. $\frac{\ln x}{\ln 2}$ and $\frac{\ln2}{\ln x}$)M1
- Correct equation in one variable, e.g. $u-\dfrac{3}{u}=2$ where $u=\log_2x$A1
- Forms and attempts to solve a quadratic: $u^{2}-2u-3=0\Rightarrow(u-3)(u+1)=0$M1
- $\log_2 x=3$ or $\log_2x=-1$A1
- $x=8$A1
- $x=\dfrac{1}{2}$ (accept $0.5$, $2^{-1}$)A1
- Answer: $x=8$ or $x=\dfrac12$
- Examiner tip: Students often find $u=3$ and $u=-1$ and then reject $u=-1$; a negative logarithm is fine, it just gives $x=2^{-1}$.
Consider the equation $4^{x}-k\times2^{x}+9=0$, where $k\in\mathbb{R}$.
(a) Show that, using $u=2^{x}$, the equation can be written as $u^{2}-ku+9=0$. [1]
(b) For $k=10$, solve the equation, giving your answers in exact form. [3]
(c) Find the set of values of $k$ for which the equation has two distinct real solutions for $x$. [4]
- (a) $4^{x}=(2^{2})^{x}=(2^{x})^{2}=u^{2}$ clearly shown, leading to $u^{2}-ku+9=0$ (AG)A1
- (b) Solves $u^{2}-10u+9=0$: $(u-1)(u-9)=0$M1
- $x=0$A1
- $x=\log_2 9$ (accept $2\log_23$, $\dfrac{\ln9}{\ln2}$)A1
- (c) States that $u=2^{x}>0$, so two distinct positive roots in $u$ are needed (each positive $u$ gives exactly one $x$)R1
- Uses the discriminant: $k^{2}-36>0$M1
- Uses product of roots $=9>0$ and sum of roots $=k>0$ (or equivalent argument, e.g. vertex $\frac{k}{2}>0$) to reject $k<-6$M1
- $k>6$ (accept $]6,\infty[$ or $(6,\infty)$; A0 for $k<-6$ or $k>6$)A1
- Answer: (b) $x=0$ or $x=\log_2 9$ (c) $k>6$
- Examiner tip: The discriminant alone gives $k<-6$ or $k>6$; the mark is lost by not using $2^{x}>0$ to reject $k<-6$, where both roots in $u$ are negative.
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