Indices, Surds and Logarithms, Worked Examples (KBAT)

Two logarithms with the same base are added, so combine them first with the product law $\log_{a}M+\log_{a}N=\log_{a}(MN)$. This turns the left side into a single logarithm:

Now undo the logarithm using the definition $\log_{a}N=c \Leftrightarrow N=a^{c}$. The base is $2$ and the right side is $4$, so the argument equals $2^{4}=16$.

The bracket is a difference of two squares, $(x+3)(x-3)=x^{2}-9$:

The algebra offers two values, but a logarithm is only defined for a positive argument. The original equation contains $\log_{2}(x-3)$, which needs $x-3>0$, that is $x>3$.

The value $x=-5$ fails this (it would give $\log_{2}(-8)$, which does not exist), so we reject it.

Only $x=5$ survives. Check it in the original equation: $\log_{2}(5+3)+\log_{2}(5-3)=\log_{2}8+\log_{2}2=3+1=4$, which matches.

So the solution is $x=5$.

The equation looks awkward until you notice $3^{2x}=(3^{x})^{2}$. Writing it that way shows a quadratic hiding inside.

Let $y=3^{x}$; then $3^{2x}=y^{2}$, and the equation becomes an ordinary quadratic in $y$:

Factorise: we need two numbers multiplying to $9$ and adding to $-10$, namely $-1$ and $-9$. So the quadratic factorises and gives two values of $y$:

Now return to $x$ by reversing the substitution $y=3^{x}$. Both values are positive, so both are valid (a power $3^{x}$ is always positive).

Write each as a power of $3$: $1=3^{0}$ and $9=3^{2}$.

So $x=0$ or $x=2$. Check both: for $x=0$, $3^{0}-10\cdot 3^{0}+9 = 1-10+9=0$; for $x=2$, $3^{4}-10\cdot 3^{2}+9 = 81-90+9=0$.

Both satisfy the equation.

Squaring $x$ and $\frac{1}{x}$ directly would be messy, so use the identity $\left(x+\dfrac{1}{x}\right)^{2}=x^{2}+2+\dfrac{1}{x^{2}}$. Rearranged, this gives $x^{2}+\dfrac{1}{x^{2}}=\left(x+\dfrac{1}{x}\right)^{2}-2$.

So we only need $x+\dfrac{1}{x}$ first.

Find $\dfrac{1}{x}$ by rationalising, multiplying top and bottom by the conjugate $\sqrt{3}-\sqrt{2}$. The denominator becomes $(\sqrt{3})^{2}-(\sqrt{2})^{2}=3-2=1$:

Add $x$ and $\dfrac{1}{x}$; the $\sqrt{2}$ terms cancel neatly, which is why this identity is the efficient route:

Now substitute into the rearranged identity, squaring $2\sqrt{3}$ as $(2)^{2}(\sqrt{3})^{2}=4\times 3=12$:

So $x^{2}+\dfrac{1}{x^{2}}=10$. A numerical check: $x\approx 3.146$ gives $x^{2}\approx 9.899$, and $\dfrac{1}{x^{2}}\approx 0.101$; their sum is $10.0$, confirming the exact answer.

Break $90$ into factors that connect to the given logarithms: $90=9\times2\times5$. Since $9=3^{2}$, working in base $3$ turns $\log_{3}90$ into a sum you can build entirely from $m$ and $n$.

The question asks for base $9$, not base $3$, so change base using $\log_{9}90=\dfrac{\log_{3}90}{\log_{3}9}$, and $\log_{3}9=2$.

The bases $2$ and $5$ don't match and neither is a power of the other, so take $\lg$ of both sides and bring the powers down using $\lg(a^{k})=k\lg a$.

Expand both sides and collect the $x$ terms on one side. Notice $\lg5+\lg2=\lg10=1$, and $3\lg2-\lg5=\lg8-\lg5=\lg\!\left(\dfrac{8}{5}\right)=\lg1.6$, which keeps the working clean.

Square both sides to remove the surd. Squaring can introduce extra solutions that don't actually satisfy the original equation, so every root found this way must be checked afterwards.

Rearrange into a quadratic equal to zero, then factorise.

Test both values in the original equation. At $x=0$: the left side is $\sqrt{1}=1$ but the right side is $0-1=-1$; a square root can never equal a negative number, so $x=0$ is rejected.

At $x=5$: the left side is $\sqrt{16}=4$ and the right side is $5-1=4$, which match.

Use the quotient law to combine the logarithms into one, then undo the log using its definition to get a simple relationship between $x$ and $y$.

Substitute this into the linear equation $x+y=24$ to solve for $y$, then back-substitute for $x$.

Since the population doubles every $3$ hours, model it as $N=500\times2^{t/3}$, where $t$ is the time in hours. Set $N=5000$ and simplify.

Take $\lg$ of both sides to bring the exponent down, then solve for $t$.