Indices, Surds and Logarithms, Worked Examples (medium)

The two sides have different bases, $4$ and $8$, so we cannot compare indices yet. The fix is to notice that both $4$ and $8$ are powers of $2$: $4=2^{2}$ and $8=2^{3}$.

Rewriting each side over the base $2$ will let us equate indices.

Replace $4$ and $8$, then use the power law $(a^{m})^{n}=a^{mn}$, multiply the indices when a power is raised to a power. Be careful to keep the whole exponent $x-1$ in a bracket:

The bases match, so the indices must be equal. Set $2x = 3x-3$ and solve the linear equation:

So $x=3$. Check both sides of the original equation: the left is $4^{3}=64$, and the right is $8^{3-1}=8^{2}=64$.

They agree, so $x=3$ is correct.

A surd in the denominator should be removed by rationalising. Multiply top and bottom by the conjugate of the denominator, the same two terms with the sign between them flipped.

Here the denominator is $\sqrt{5}-\sqrt{2}$, so its conjugate is $\sqrt{5}+\sqrt{2}$. Multiplying by $\dfrac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}$ is multiplying by $1$, so the value is unchanged.

The denominator is a difference of two squares, $(a-b)(a+b)=a^{2}-b^{2}$, which is exactly why the conjugate works, the surds disappear:

Now the expression is $\dfrac{6(\sqrt{5}+\sqrt{2})}{3}$. Divide the numerical factor $6$ by $3$ to get $2$:

So $a=2$ and $b=2$, giving $2\sqrt{5}+2\sqrt{2}$. A numerical check: the original is $\dfrac{6}{2.236-1.414}\approx\dfrac{6}{0.822}\approx7.30$, and $2\sqrt{5}+2\sqrt{2}\approx4.47+2.83=7.30$, so the answer is confirmed.

We are only given the logarithms of $2$ and $3$, so the plan is to break $24$ into factors made only of $2$s and $3$s. Prime-factorise: $24 = 8\times 3 = 2^{3}\times 3$.

Writing $24$ this way lets the log laws convert it into $\log_{a}2$ and $\log_{a}3$.

Apply the product law $\log_{a}(MN)=\log_{a}M+\log_{a}N$ to split the factors:

Now use the power law $\log_{a}M^{k}=k\log_{a}M$ on the first term to bring the index $3$ down in front:

Finally substitute the given values $\log_{a}2=p$ and $\log_{a}3=q$:

So $\log_{a}24 = 3p+q$. The check is structural: $24=2^{3}\times 3$ uses three factors of $2$ (giving $3p$) and one factor of $3$ (giving $q$), which matches the answer exactly.

None of these three surds match yet, so before we can add or subtract we must write each one as a whole number times $\sqrt{3}$. Break the number under each root into a perfect-square factor times $3$.

Once every term is a multiple of the same surd $\sqrt{3}$, they behave like like terms in algebra, add or subtract the whole-number coefficients only.

A single logarithm equal to a number converts directly to index form: $\log_a N = m$ means $N=a^{m}$. Here the base is $3$ and the logarithm equals $2$, so rewrite the equation without the logarithm.

Simplify the right side, $3^{2}=9$, then solve the resulting linear equation for $x$.

The equation is quadratic in form once we notice $2^{2x}=(2^{x})^{2}$. Let $y=2^{x}$, where $y>0$; the equation becomes an ordinary quadratic in $y$.

Factorise the quadratic, solve for $y$, then convert each positive value of $y$ back to $2^{x}$ and read off $x$.

The two logarithms use related bases, since $4=2^{2}$. Rewrite the given logarithm with base $2$ using $\log_{a^{n}}N=\dfrac{1}{n}\log_{a}N$, which halves the logarithm when the base is squared.

Substitute the given value $\log_{4}y=3$ and solve for $\log_{2}y$.

A negative index means 'take the reciprocal', so first flip the fraction inside the brackets and change the sign of the index to positive.

Recognise $81$ and $16$ as fourth powers, $81=3^{4}$ and $16=2^{4}$, so the base is already a fourth power. A fractional index $\tfrac34$ means 'fourth root, then cube' (or the other order).