E3.7 - Logarithms and exponential equations (Extended)

E3.7 - Logarithms and exponential equations (Extended)

How long does an investment take to grow by half when the same percentage is added each year? E3.7 introduces logarithms as the inverse of exponentials, then uses them to find unknown powers and interpret growth and decay times. Graphs provide a second way to see and check the solutions.

Inverse of exponentials

You know that 25=322^5=32. If the missing number is the power, the question becomes “What power of 2 gives 32?” A logarithm records that power: log⁡232=5\log_2 32=5.

Logarithm

For a>0a>0, a≠1a\ne1 and y>0y>0,

y=ax⟺x=log⁡ayy=a^x\quad\Longleftrightarrow\quad x=\log_a y

The small number aa is the base. The logarithm is the exponent, not the result of division by the base. A logarithm written log⁡y\log y has base 10 unless another base is stated.

The exponential function sends an input xx to axa^x; its inverse sends that positive output back to xx. For example, 100=110^0=1 gives log⁡1=0\log1=0, and 10−2=0.0110^{-2}=0.01 gives log⁡0.01=−2\log0.01=-2. A logarithm can be negative even though its argument must be positive. No real logarithm of zero or a negative number is defined here.

[DIAGRAM: asset_name: Exponential and logarithm reverse input and output; asset_slug: logarithm_inverse; caption: Corresponding points swap coordinates across the line y equals x.; description: Two accurate curves y equals two to the power x and y equals log base two of x, reflected in the dashed line y equals x. The pairs zero comma one and one comma zero, one comma two and two comma one, and two comma four and four comma two are marked. The logarithm exists only to the right of the vertical axis.]
Corresponding points swap coordinates across the line y equals x.

Worked example 1: reading a power backwards

Question. Find log⁡232\log_2 32.

Plan. Find the exponent that makes a power of 2 equal 32.

25=32⇒log⁡232=52^5=32\quad\Rightarrow\quad\log_2 32=5

Answer. 5.

Check. Five factors of 2 multiply to 32, so the inverse returns the original exponent.

A base-ten logarithm can contain an expression. Reverse the logarithm first, then solve the resulting ordinary equation.

Worked example 2: an expression inside a logarithm

Question. Solve log⁡(4x)=2\log(4x)=2.

Plan. Rewrite the logarithmic equation as an exponential equation with base 10.

4x=102reverse the logarithmx=100÷4=25undo multiplication\begin{aligned} 4x&=10^2 &&\text{reverse the logarithm}\\ x&=100\div4=25 &&\text{undo multiplication} \end{aligned}

Answer. x=25x=25.

Check. 4(25)=100>04(25)=100>0, so the logarithm is defined, and log⁡100=2\log100=2. The answer is not 102/log⁡410^2/\log4: the argument is the whole product 4x4x.

This method uses the inverse relationship directly; no logarithm laws are needed.

Watch out

log⁡(4x)=2\log(4x)=2 means 4x=1024x=10^2, not 4x=2104x=2^{10}. The unstated base belongs to the exponential, while the logarithm's value becomes the exponent.

Reverse the function before dividing in this fresh equation.

Solving exponentials

Not every exponential equation has an integer solution. Since 32=93^2=9 and 33=273^3=27, the solution of 3x=203^x=20 lies between 2 and 3. A calculator's base-ten logarithms give it directly.

Finding an unknown exponent

If au=ba^u=b, where a>0a>0, a≠1a\ne1 and b>0b>0, then

u=log⁡blog⁡au=\frac{\log b}{\log a}

Use the same logarithm base in numerator and denominator. This relationship is not on the supplied formula list for Papers 1–4: learn it.

The quotient changes the base of the logarithm: it evaluates log⁡ab\log_a b using the calculator's ordinary log key. Keep the whole quotient together. When an expression occupies the exponent, the quotient equals that entire expression.

Worked example 3: a non-integer exponent

Question. Solve 3x=203^x=20, giving your answer to 3 significant figures.

Plan. Use the logarithm quotient, then check the answer between neighbouring powers.

x=log⁡20log⁡3=2.726833…x=\frac{\log20}{\log3}=2.726833\ldots

Answer. x=2.73x=2.73 to 3 significant figures.

Check. The unrounded value gives 32.726833…=203^{2.726833\ldots}=20. Also 9<20<279<20<27, so 2<x<32<x<3, consistent with the answer.

Only the exponent changes in the next equation. Find the whole exponent before isolating xx.

Worked example 4: an expression as the exponent

Question. Solve 32x−1=203^{2x-1}=20, to 3 significant figures.

Plan. Set 2x−12x-1 equal to the same logarithm quotient, then rearrange.

2x−1=log⁡20log⁡3=2.726833…x=2.726833…+12=1.863416…\begin{aligned} 2x-1&=\frac{\log20}{\log3}=2.726833\ldots\\ x&=\frac{2.726833\ldots+1}{2}=1.863416\ldots \end{aligned}

Answer. x=1.86x=1.86 to 3 significant figures.

Check. Using the unrounded xx, 2x−1=2.726833…2x-1=2.726833\ldots, and raising 3 to that power returns 20. Substituting x=2.73x=2.73 instead would put the wrong number in the exponent.

You can also plot the exponential and the horizontal line on a GDC, choose a window containing their crossing and use the permitted intersection feature. Record the graphs and the intersection when a graphical method is requested; an unlisted equation-solver application is not the graphical method.

Watch out

Do not reverse the quotient: log⁡3/log⁡20\log3/\log20 is less than 1 and cannot be the power producing 20 from 3. Keep full calculator precision until the final answer.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account