### Exponentials & Logarithms

Problem requiring decisions

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## Solution

Some equations involving powers or indices can be solved using logarithms… but not all.

Think about how you could go about solving the following equations. Sort them according to the tools or methods you would use.

(A) $3^x=81$

We know that $81=9\times9=3^4$ so we can say that $x=4$ using only our knowledge of powers of $3$. Note that we could instead use logarithms to solve this equation, \begin{align*} \log_3{3^x}&=\log_3{81}\\ x&=\log_3{81} \end{align*}

so we could classify this equation as “either indices or logarithms”.

(B) $x^5=50$

To solve this equation, remember that the inverse of raising to a power is a root or a fractional power. So we can solve it using indices, $x=50^\frac{1}{5}=\sqrt[5]{50} \text{ .}$

(C) $3^x=43$

This looks very similar to the first equation, but since $43$ is not an exact power of $3$ we have to use logarithms if we want an exact solution.

Have a go at estimating the solution.

(D) $5^{2x}-5^x-6=0$

The laws of indices tell us that $5^{2x}=(5^x)^2$ so this is a quadratic equation which we can factorise \begin{align*} (5^x)^2-(5^x)-6&=0\\ (5^x-3)(5^x+2)&=0\\ 5^x&=3 \text{ or } -2 \text{ .} \end{align*}

Now, $5^x$ cannot be negative so we have only one solution, for which we shall need logarithms. $x=\log_5 3$ So we could classify this equation as requiring rearrangement and logarithms.

(E) $5^x+4^x=8$

We don’t know a way to solve this exactly. How could we find an approximate solution?
What’s wrong with this? \begin{align*} \log{(5^x+4^x)}&=\log{8}\\ \log{5^x}+\log{4^x}&=\log{8}\\ x\log5+x\log4&=\log8\\ x&=\frac{\log8}{\log5+\log4} \end{align*}

(F) $5^x+2\times5^{1-x}=7$

It may not be immediately obvious what kind of equation this is, so we want to manipulate it into a more useful form. What can we do to the equation so there is no $-x$ in the index of the second term?

(G) $3^{2x}-3=24$

This could be treated as a quadratic in $3^x$, but a bit of rearrangement reveals a simpler method to solve it. \begin{align*} 3^{2x}&=27\\ 2x&=3 \end{align*}

It’s always worth looking for simpler methods before embarking on complicated ones. In this case, gathering the constant terms together was the inviting first step.

(H) $2^{2x}-9\times2^x+8=0$

This is another quadratic, this time in $2^x$.

(I) $\sqrt{2x-3}=5$

Logarithms are not going to help, but squaring both sides will.

(J) $5^x-x^5=3$

This is another example of an equation we don’t know how to solve exactly. We could use a numerical method such as trial and improvement to find an approximate solution.

How many solutions are there? Most numerical methods work best if you know roughly where the solution is. A sketch of $y=5^x$ and $y=x^5$ might help.

(K) $16^{\frac{3}{x}}=8$

What’s gone wrong here? \begin{align*} 16^{\frac{3}{x}}&=8\\ 2^{\frac{3}{x}}&=1\\ \frac{2^3}{2^x}&=1\\ 2^3&=2^x\\ x&=3 \end{align*}

Have the two methods given the same answer? Which method do you prefer?

(L) $\big(\frac{13}{16}\big)^{3x}=\frac{3}{4}$

We are going to have to use logarithms on this one.

Why would indices not give us a solution?

There is a fundamental difference between equations involving the unknown, $x$, raised to a fixed power (such as $x^5$) and equations involving a constant raised to an unknown power (such as $5^x$). How is this difference reflected in the methods required to solve the equations?