2.10 Exponential Equations
1. Solving Variable Exponential Systems
An unknown variable mapping within the mathematical exponent utilizes logarithms inherently for absolute isolation. The purest format $a^x = b$ evaluates accurately to $x = \log_a b$ or $x = \dfrac{\ln b}{\ln a}$.
EXAMPLE 1
Solve $2(5^x) = 9$.
Isolate strictly by dividing by 2: $5^x = 4.5$.
Extract algebraically using logs:
$\log(5^x) = \log(4.5) \Rightarrow x\log 5 = \log 4.5 \Rightarrow \mathbf{x = \dfrac{\log 4.5}{\log 5}}$.
$\log(5^x) = \log(4.5) \Rightarrow x\log 5 = \log 4.5 \Rightarrow \mathbf{x = \dfrac{\log 4.5}{\log 5}}$.
Notice: If we use $\ln()$, the answer will be $x = \dfrac{\ln 4.5}{\ln 5}$. If we use $\log_5()$, the answer will be $x = \log_5 4.5$.
EXAMPLE 2
Solve $10e^{2x} = 85$.
Isolate strictly: $e^{2x} = 8.5$.
Natural logarithm usage removes base $e$:
$\ln(e^{2x}) = \ln(8.5) \Rightarrow 2x = \ln(8.5) \Rightarrow \mathbf{x = \dfrac{\ln 8.5}{2}}$.
$\ln(e^{2x}) = \ln(8.5) \Rightarrow 2x = \ln(8.5) \Rightarrow \mathbf{x = \dfrac{\ln 8.5}{2}}$.
Notice: Whenever we see exponentials of base $e$, it is preferable to use $\ln()$.
EXAMPLE 3
Solve $5^x = 2^{x+1}$ cleanly and express via the format $\dfrac{\ln a}{\ln b}$.
Method A: Apply natural log directly.
$5^x = 2^{x+1} \Rightarrow \ln(5^x) = \ln(2^{x+1})$
$x\ln 5 = (x+1)\ln 2 \Rightarrow x\ln 5 = x\ln 2 + \ln 2$.
Isolate components containing $x$: $x(\ln 5 - \ln 2) = \ln 2 \Rightarrow \mathbf{x = \dfrac{\ln 2}{\ln(5/2)}}$.
$5^x = 2^{x+1} \Rightarrow \ln(5^x) = \ln(2^{x+1})$
$x\ln 5 = (x+1)\ln 2 \Rightarrow x\ln 5 = x\ln 2 + \ln 2$.
Isolate components containing $x$: $x(\ln 5 - \ln 2) = \ln 2 \Rightarrow \mathbf{x = \dfrac{\ln 2}{\ln(5/2)}}$.
Method B: Combine algebraically first.
$5^x = 2^x \cdot 2 \Rightarrow \dfrac{5^x}{2^x} = 2 \Rightarrow \left(\dfrac{5}{2}\right)^x = 2$.
Apply natural log: $x\ln\left(\dfrac{5}{2}\right) = \ln 2 \Rightarrow \mathbf{x = \dfrac{\ln 2}{\ln(5/2)}}$.
$5^x = 2^x \cdot 2 \Rightarrow \dfrac{5^x}{2^x} = 2 \Rightarrow \left(\dfrac{5}{2}\right)^x = 2$.
Apply natural log: $x\ln\left(\dfrac{5}{2}\right) = \ln 2 \Rightarrow \mathbf{x = \dfrac{\ln 2}{\ln(5/2)}}$.
Remarks: This is the exact answer. If an expression in the form $\log_a b$ is required, the answer is $x = \log_{\frac{5}{2}} 2$.
2. Exponential Modelling (Growth or Decay)
In many applications, a quantity increases or decreases exponentially according to time. Suppose that a population $P$ at time $t$ is given by the formula $P = P_0 e^{kt}$.
- $P_0$ represents unequivocally the initial evaluated starting parameter at mathematically established time zero ($t=0$).
- A positive constant $k > 0$ models exponential runaway growth (e.g., $P = 1000e^{0.2t}$).
- A negative constant $k < 0$ models collapsing exponential decay (e.g., $P = 1000e^{-0.2t}$).
EXAMPLE 4 (Growth Applications)
Consider the case where the initial population is 1000 and $P$ is given by $P = 1000e^{0.2t}$.
Question 1: What is the initial population (at starting time)?
Initial means $t=0$. Since $e^0 = 1$, $P = \mathbf{1000}$.
Initial means $t=0$. Since $e^0 = 1$, $P = \mathbf{1000}$.
Question 2: What is the population after 3 years?
For $t=3$, $P = 1000e^{0.2 \times 3} = 1000e^{0.6} \approx \mathbf{1822}$.
For $t=3$, $P = 1000e^{0.2 \times 3} = 1000e^{0.6} \approx \mathbf{1822}$.
Question 3: The population after $t$ years is 2500. Find $t$.
$1000e^{0.2t} = 2500 \Rightarrow e^{0.2t} = 2.5 \Rightarrow \ln(e^{0.2t}) = \ln(2.5) \Rightarrow 0.2t = \ln(2.5) \Rightarrow t = \dfrac{\ln 2.5}{0.2} \approx \mathbf{4.58 \text{ years}}$.
$1000e^{0.2t} = 2500 \Rightarrow e^{0.2t} = 2.5 \Rightarrow \ln(e^{0.2t}) = \ln(2.5) \Rightarrow 0.2t = \ln(2.5) \Rightarrow t = \dfrac{\ln 2.5}{0.2} \approx \mathbf{4.58 \text{ years}}$.
Question 4: The population doubles after $t$ years. Find $t$.
We set $P = 2000$. $1000e^{0.2t} = 2000 \Rightarrow e^{0.2t} = 2 \Rightarrow 0.2t = \ln 2 \Rightarrow t = \dfrac{\ln 2}{0.2} \approx \mathbf{3.47 \text{ years}}$.
We set $P = 2000$. $1000e^{0.2t} = 2000 \Rightarrow e^{0.2t} = 2 \Rightarrow 0.2t = \ln 2 \Rightarrow t = \dfrac{\ln 2}{0.2} \approx \mathbf{3.47 \text{ years}}$.
Question 5 (Finding k): Suppose $P = P_0 e^{kt}$. Given that the population doubles every 4 years, find $k$.
For $t=4$, $P = 2P_0$. Thus $P_0 e^{4k} = 2P_0 \Rightarrow e^{4k} = 2 \Rightarrow 4k = \ln 2 \Rightarrow k = \dfrac{\ln 2}{4} \approx \mathbf{0.173}$.
For $t=4$, $P = 2P_0$. Thus $P_0 e^{4k} = 2P_0 \Rightarrow e^{4k} = 2 \Rightarrow 4k = \ln 2 \Rightarrow k = \dfrac{\ln 2}{4} \approx \mathbf{0.173}$.
EXAMPLE 5 (Decay Applications)
The mass $m$ of a radio-active substance at time $t$ hours is given by $m = 4e^{-kt}$.
(a) The mass is 1 kg after 5 hours. Find $k$.
For $t=5$, $m=1$, thus $4e^{-5k} = 1 \Rightarrow e^{-5k} = \dfrac{1}{4} \Rightarrow -5k = \ln(0.25) \Rightarrow k = \dfrac{\ln(0.25)}{-5} \approx \mathbf{0.277}$.
Therefore, the model maps exactly to $m = 4e^{-0.277t}$.
For $t=5$, $m=1$, thus $4e^{-5k} = 1 \Rightarrow e^{-5k} = \dfrac{1}{4} \Rightarrow -5k = \ln(0.25) \Rightarrow k = \dfrac{\ln(0.25)}{-5} \approx \mathbf{0.277}$.
Therefore, the model maps exactly to $m = 4e^{-0.277t}$.
(b) What is the mass after 3 hours?
For $t=3$, evaluating physical mass parameters gives: $m = 4e^{-0.277 \times 3} \approx \mathbf{1.74 \text{ kg}}$.
For $t=3$, evaluating physical mass parameters gives: $m = 4e^{-0.277 \times 3} \approx \mathbf{1.74 \text{ kg}}$.
(c) The mass reduces to a half after $t$ hours. Find $t$.
For $m=2$, $4e^{-0.277t} = 2 \Rightarrow e^{-0.277t} = 0.5 \Rightarrow -0.277t = \ln(0.5) \Rightarrow t = \dfrac{\ln(0.5)}{-0.277} \approx \mathbf{2.50 \text{ hours}}$.
(This specific time is scientifically known as the half-life time).
For $m=2$, $4e^{-0.277t} = 2 \Rightarrow e^{-0.277t} = 0.5 \Rightarrow -0.277t = \ln(0.5) \Rightarrow t = \dfrac{\ln(0.5)}{-0.277} \approx \mathbf{2.50 \text{ hours}}$.
(This specific time is scientifically known as the half-life time).
3. More Exponential Equations (HL)
Common Mistake Warning: The statement $A \pm B = C$ does not imply $\log A \pm \log B = \log C$. It implies $\log(A \pm B) = \log C$. If an equation contains a sum of exponentials, it doesn't help to apply a logarithm, as $\log(a^x \pm b^x)$ cannot be mathematically simplified. In such non-linear polynomial equations, we usually substitute an exponential with a new variable $y$.
EXAMPLE 6
Solve $6^x 7^{x-1} = 3^{x-2}$.
Separate indices algebraically: $\dfrac{6^x 7^x}{7} = \dfrac{3^x}{3^2}$.
Regroup entirely corresponding operational base variables:
$\dfrac{42^x}{3^x} = \dfrac{7}{9} \Rightarrow 14^x = \dfrac{7}{9}$.
$\dfrac{42^x}{3^x} = \dfrac{7}{9} \Rightarrow 14^x = \dfrac{7}{9}$.
Apply log functions explicitly: $x\ln 14 = \ln\left(\dfrac{7}{9}\right) \Rightarrow \mathbf{x = \dfrac{\ln(7/9)}{\ln 14}}$.
EXAMPLE 7 (Quadratic Substitution)
(a) Solve $6e^x + \dfrac{12}{e^x} = 17$.
Substitute carefully mapped values $y = e^x$:
$6y + \dfrac{12}{y} = 17 \Rightarrow 6y^2 - 17y + 12 = 0$.
Factoring quadratic elements generates physical solutions $y = \dfrac{3}{2}$ or $y = \dfrac{4}{3}$.
Reversing initial parameter mapping yields completely verified independent valid targets:
$e^x = \dfrac{3}{2} \Rightarrow \mathbf{x = \ln\left(\dfrac{3}{2}\right)}$, and $e^x = \dfrac{4}{3} \Rightarrow \mathbf{x = \ln\left(\dfrac{4}{3}\right)}$.
$6y + \dfrac{12}{y} = 17 \Rightarrow 6y^2 - 17y + 12 = 0$.
Factoring quadratic elements generates physical solutions $y = \dfrac{3}{2}$ or $y = \dfrac{4}{3}$.
Reversing initial parameter mapping yields completely verified independent valid targets:
$e^x = \dfrac{3}{2} \Rightarrow \mathbf{x = \ln\left(\dfrac{3}{2}\right)}$, and $e^x = \dfrac{4}{3} \Rightarrow \mathbf{x = \ln\left(\dfrac{4}{3}\right)}$.
(b) Solve $6(10^{2x}) + 12 = 17(10^x)$.
Substitute $y = 10^x$. This transforms the equation into: $6y^2 - 17y + 12 = 0$.
The solutions are again $y = \dfrac{3}{2}$ or $y = \dfrac{4}{3}$.
Reversing the mapping yields:
$10^x = \dfrac{3}{2} \Rightarrow \mathbf{x = \log\left(\dfrac{3}{2}\right)}$, and $10^x = \dfrac{4}{3} \Rightarrow \mathbf{x = \log\left(\dfrac{4}{3}\right)}$.
The solutions are again $y = \dfrac{3}{2}$ or $y = \dfrac{4}{3}$.
Reversing the mapping yields:
$10^x = \dfrac{3}{2} \Rightarrow \mathbf{x = \log\left(\dfrac{3}{2}\right)}$, and $10^x = \dfrac{4}{3} \Rightarrow \mathbf{x = \log\left(\dfrac{4}{3}\right)}$.
EXAMPLE 8 (Systems)
Solve the multi-variable algebraic system:
$2(3^x) - 3(2^y) = -22$
$5(3^x) + \dfrac{1}{2}(2^y) = 9$
Substitute variables: Let $A = 3^x$ and $B = 2^y$.
The linear algebraic system immediately translates functionally to:
$2A - 3B = -22$
$5A + 0.5B = 9$
$2A - 3B = -22$
$5A + 0.5B = 9$
The solved linear output accurately returns independent components $A = 1$ and $B = 8$.
Reverse exponential equations yield independent numerical inputs:
$3^x = 1 \Rightarrow \mathbf{x=0}$
$2^y = 8 \Rightarrow \mathbf{y=3}$.
$3^x = 1 \Rightarrow \mathbf{x=0}$
$2^y = 8 \Rightarrow \mathbf{y=3}$.