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 $x = \frac{\ln b}{\ln a}$.

EXAMPLE 1 & 2

Example 1: Solve $2(5^x) = 9$.
Isolate strictly: $5^x = 4.5$.
Extract algebraically: $\ln(5^x) = \ln(4.5) \Rightarrow x\ln 5 = \ln 4.5 \Rightarrow \mathbf{x = \frac{\ln 4.5}{\ln 5}}$.
Example 2: Solve $10e^{2x} = 85$.
Isolate strictly: $e^{2x} = 8.5$.
Natural logarithm usage removes base $e$: $\ln(e^{2x}) = \ln(8.5) \Rightarrow 2x = \ln(8.5) \Rightarrow \mathbf{x = \frac{\ln 8.5}{2}}$.

EXAMPLE 3

Solve $5^x = 2^{x+1}$ cleanly and express via the format $\frac{\ln a}{\ln b}$.

Method A: Apply natural log directly.
$x\ln 5 = (x+1)\ln 2 \Rightarrow x\ln 5 = x\ln 2 + \ln 2$.
Isolate components containing $x$: $x(\ln 5 - \ln 2) = \ln 2 \Rightarrow \mathbf{x = \frac{\ln 2}{\ln(5/2)}}$.
Method B: Combine algebraically first.
$5^x = 2^x \cdot 2 \Rightarrow \frac{5^x}{2^x} = 2 \Rightarrow \left(\frac{5}{2}\right)^x = 2$.
Apply natural log: $x\ln\left(\frac{5}{2}\right) = \ln 2 \Rightarrow \mathbf{x = \frac{\ln 2}{\ln(5/2)}}$.

2. Exponential Modelling (Growth or Decay)

Real-world biological and physical quantities model universally using the continuous equation $P = P_0 e^{kt}$.

  • $P_0$ represents unequivocally the initial evaluated starting parameter at mathematically established time zero ($t=0$).
  • A positive constant $k > 0$ models exponential runaway growth.
  • A negative constant $k < 0$ models collapsing exponential decay (half-life calculations).

EXAMPLE 4 (Decay Applications)

A mass function mapping a specific radio-active decay tracks $m = 4e^{-kt}$.

(a) Given the physical mass scales exactly to 1kg strictly after 5 hours, evaluate $k$.
$4e^{-5k} = 1 \Rightarrow e^{-5k} = 0.25 \Rightarrow -5k = \ln(0.25) \Rightarrow k = -\frac{\ln(0.25)}{5} \approx \mathbf{0.277}$.
(b) Evaluating physical mass parameters at precisely 3 hours:
$m = 4e^{-0.277 \times 3} \approx \mathbf{1.74}$.
(c) Calculate strict half-life time required to collapse mass to exactly $m=2$.
$4e^{-0.277t} = 2 \Rightarrow e^{-0.277t} = 0.5 \Rightarrow -0.277t = \ln(0.5) \Rightarrow t = \frac{\ln(0.5)}{-0.277} \approx \mathbf{2.50 \text{ hours}}$.

3. More Exponential Equations (HL)

Non-linear polynomial expressions masked as exponentials resolve utilizing a strict substitution process (setting $y = e^x$).

EXAMPLE 5

Solve $6^x 7^{x-1} = 3^{x-2}$.

Separate indices algebraically: $\frac{6^x 7^x}{7} = \frac{3^x}{3^2}$.
Regroup entirely corresponding operational base variables: $\frac{42^x}{3^x} = \frac{7}{9} \Rightarrow 14^x = \frac{7}{9}$.
Apply log functions explicitly: $x\ln 14 = \ln(7/9) \Rightarrow \mathbf{x = \frac{\ln(7/9)}{\ln 14}}$.

EXAMPLE 6 (Quadratic Substitution)

Solve $6e^x + \frac{12}{e^x} = 17$.

Substitute carefully mapped values $y = e^x$: $6y + \frac{12}{y} = 17 \Rightarrow 6y^2 - 17y + 12 = 0$.
Factoring quadratic elements generates physical solutions $y = 3/2$ or $y = 4/3$.
Reversing initial parameter mapping yields completely verified independent valid targets: $e^x = 3/2 \Rightarrow \mathbf{x = \ln(3/2)}$, and $e^x = 4/3 \Rightarrow \mathbf{x = \ln(4/3)}$.

EXAMPLE 7 (Systems)

Solve the multi-variable algebraic system:
$2(3^x) - 3(2^y) = -22$
$5(3^x) + \frac{1}{2}(2^y) = 9$

Substitute $A = 3^x$ and $B = 2^y$.
The linear algebraic system immediately translates functionally to:
$2A - 3B = -22$
$5A + 0.5B = 9$
The solved linear output accurately returns independent components $A = 1, B = 8$.
Reverse exponential equations yield independent numerical inputs: $3^x = 1 \Rightarrow \mathbf{x=0}$ and $2^y = 8 \Rightarrow \mathbf{y=3}$.