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Equations Involving Exponentials
Equations Involving Exponentials
Equations involving exponentials and logarithms can become much more complicated than we have already seen. On this page, we will learn how to use a calculator for logarithms, attempt to solve some more difficult equations, and apply knowledge of logarithms to real life.
The following topics build on the content in this page.
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View ProductUsing a Calculator
Your calculator will have several buttons to do with logarithms.
\log_{\square}\square allows you to do any logarithm. Put the base in the lower box and the number in the box on the right.
\log_{10}\square is for logarithms with a base of 10 only.
\ln is the natural logarithm. We will see this later.
Example: If we want to do \log_{4}(81) on a calculator we would press \log_{\square}\square then 4 then 8 then 1 then =
Exponential Equations
Some exponential equations look more complicated, but they are really just quadratics in disguise.
They take the form ap^{2x}+bp^{x}+c=0
We solve them by substituting y=p^{x}, and noticing that y^{2}=p^{2x}
Then we have a simple quadratic: ay^{2}+by+c=0
This gives us two roots: y=r_{1} and y=r_{2}
Putting x back in gives p^{x}=r_{1} and p^{x}=r_{2}, which are two equations that we already know how to solve.
Example: 3\times6^{2x}-7\times6^{x}+2=0
3\times(6^{x})^{2}-7\times6^{x}+2=0
y=6^{x}
3y^{2}-7y-2=0
(3y-1)(y-2)=0
y=\dfrac{1}{3} or y=2
6^{x}=\dfrac{1}{3} or 6^{x}=2
x=\log_{6}\left(\dfrac{1}{3}\right) or x=\log_{6}(2)
x=-0.613 or x=0.387
Logarithm Equations
Difficult logarithm equations require you to use more than one law of logarithms.
Example: \log_{3}(21x-2)-2\log_{3}(x)=3
\log_{3}(21x-2)+\log_{3}(x^{-2})=3
\log_{3}((21x-2)x^{-2})=3
\log_{3}\left(\dfrac{21x-2}{x^{2}}\right)=3
\dfrac{21x-2}{x^{2}}=3^{3}
\dfrac{21x-2}{x^{2}}=27
21x-2=27x^{2}
27x^{2}-21x+2=0
(9x-1)(3x-2)=0
x=\dfrac{1}{9} and x=\dfrac{2}{3}
Real-Life Problems
Exponentials and logarithms appear frequently in real life. An important skill is being able to solve real world problems involving exponentials and logarithms.
Example: A car depreciates in value over the course of several years according to the function V=14000\times0.9^{T} where V is the value of the car and T is the time in years.
i) How much does the car cost new?
ii) Jon has owned his car for 9 years. How much is it worth?
iii) Clarissa buys a new car today. How many years until it is worth half of what she paid?
i) New is at T=0 so V=14000\times0.9^{0}=14000
ii) This is T=9 so V=14000\times0.9^{9}=5424
iii) She paid new cost which is 14000
Half of what she paid is 7000
7000=14000\times0.9^{T}
0.9^{T}=\dfrac{7000}{14000}
0.9^{T}=0.5
\begin{aligned}T&=\log_{0.9}(0.5)\\[1.2em]&=6.58\end{aligned}
Equations Involving Exponentials Example Questions
Question 1: Solve for x:
3^{2x}-28\times3^{x}+27=0[3 marks]
3^{2x}-28\times3^{x}+27=0
(3^{x})^{2}-28\times3^{x}+27=0
Make a substitution y=3^{x}
y^{2}-28y+27=0
(y-27)(y-1)=0
y=27 or y=1
Put our substitution back in:
3^{x}=27 or 3^{x}=1
x=\log_{3}(27) or x=\log_{3}(1)
x=3 or x=0
Question 2: Solve for x:
2\log_{2}(2x-1)-3\log_{2}(x)=4[5 marks]
2\log_{2}(2x-1)-3\log_{2}(x)=4
\log_{2}((2x-1)^{2})-\log_{2}(x^{3})=4
\log_{2}\left(\dfrac{(2x-1)^{2}}{x^{3}}\right)=4
\dfrac{(2x-1)^{2}}{x^{3}}=2^{4}
\dfrac{(2x-1)^{2}}{x^{3}}=16
(2x-1)^{2}=16x^{3}
4x^{2}-4x+1=16x^{3}
16x^{3}-4x^{2}+4x-1=0
(4x-1)(4x^{2}+1)=0
x=\dfrac{1}{4} is the only real solution.
Question 3: A particle’s radioactivity decreases according to R=1000\times0.8^{T} where R is radioactivity and T is time in years.
a) Find the half-life (when radioactivity has halved from the initial value) of the particle.
b) A safe level of radioactivity is 5. How long until the particle is safe?
[5 marks]
Find initial value, which is at T=0:
\begin{aligned}R&=1000\times0.8^{0}\\[1.2em]&=1000\times1\\[1.2em]&=1000\end{aligned}
Hence, half-life is when R=\dfrac{1000}{2}=500
500=1000\times0.8^{T}
\dfrac{500}{1000}=0.8^{T}
0.5=0.8^{T}
\begin{aligned}T&=\log_{0.8}(0.5)\\[1.2em]&=3.11\text{ years}\end{aligned}
R=5
5=1000\times0.8^{T}
\dfrac{5}{1000}=0.8^{T}
0.005=0.8^{T}
\begin{aligned}T&=\log_{0.8}(0.005)\\[1.2em]&=23.7\text{ years}\end{aligned}
Specification Points Covered
F5 – Solve equations of the form a^x=b
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