03Cexponentialeqns

toc = Exponential Equations =

An ** exponential equation ** (or ** indicial equation **) is an equation where the variable is in the __exponent__ (or index).

Exponential equations can be solved by expressing each side with the same base, then equating the indices.

**Example 1**
Solve the following: math \\ . \qquad a) \quad 3^{x-1} = 81 \\ . \\ . \qquad b) \quad 4^{2x-1} = \dfrac{1}{256} math

__Solution:__

math \\ . \qquad 3^{x-1}=81 \\. \\ . \qquad 3^{x-1}=3^4 \qquad \textit{equate the indices}\\. \\ . \qquad x-1=4 \\. \\ . \qquad x=5 math
 * (a) **

math \\ . \qquad 4^{2x-1}=\dfrac{1}{256} \\. \\ . \qquad 4^{2x-1}=\dfrac{1}{4^4} \\. \\ . \qquad 4^{2x-1}=4^{-4} \qquad \textit{equate the indices} math . math \\ . \qquad 2x-1=-4 \\. \\ . \qquad 2x=-3 \\. \\ . \qquad x=-\dfrac{3}{2} math
 * (b) **

**Example 2**
math . \qquad \text{Find x, where: } \;\; 3^{2x}-3^{x+2}+8=0 math

__Solution:__

math \\ . \qquad 3^{2x}-3^{x+2}+8=0 \\. \\ . \qquad \big( 3^x \big)^2 - 3^x \times 3^2 + 8 = 0 math

Let u = 3 x so equation can be written as a quadratic math \\ . \qquad u^2-9u+8=0 \\. \\ . \qquad (u-1)(u-8)=0 \\. \\ . \qquad u=1 \;\; \textit{ or } \;\; u=8 math

substitute u = 3 x math \\ . \qquad 3^x=1 \;\; \textit{ or } \;\; 3^x=8 \\. \\ . \qquad x=0 \;\;\textit{ or } \;\; x= \log_38 \; \; \{\text{ exact answer} \} math

More Exponential Equations
Sometimes exponential equations can be solved by taking the log of both sides.

**Example 3**
... Solve 3 2x = 28. ... ... ... a) in exact form using base 10 ... ... ... b) correct to 3 decimal places

__Solution:__


 * (a) **

math . \qquad 3^{2x}=28 math

Take the log(base 10) of both sides math \\ . \qquad \log_{10}3^{2x} = \log_{10}{28} \\. \\ . \qquad 2x \log_{10}3 = \log_{10}{28} math . math \\ . \qquad 2x =\dfrac{\log_{10}{28}}{\log_{10}3} \\. \\ . \\ . \qquad x = \dfrac{\log_{10}{28}}{2 \log_{10}3} \qquad \textit{exact answer} math

math . \qquad x = 1.517 \qquad \textit{correct to 3 decimal places} math
 * (b) **

This can be calculated using the ClassPad, either from the final answer in part (//**a)**//, or by solving the initial equation, as dislayed above.

**Example 4**
... Solve the following, answering in both exact form and correct to 3 decimal places: math \\ . \qquad (a) \quad 5^{3x} > 29 \\. \\ . \qquad (b) \quad 0.5^x \leqslant 8 math

__Solution:__

**(a)** math . \qquad 5^{3x} > 29 math

Take log (base 10) of both sides {or use log(base e), see below} math \\ . \qquad \log_{10}{5^{3x}} > \log_{10}{29} \\. \\ . \qquad 3x \log_{10}5 > \log_{10}{29} math . math . \qquad 3x > \dfrac{ \log_{10}{29}}{\log_{10}5} math . math . \qquad x > \dfrac{\log_{10}{29}}{3\log_{10}5} \qquad \textit{exact answer} math . math . \qquad x > 0.697 \qquad \textit{correct to 3 decimal places} math

**(b)** math \\ . \qquad 0.5^x \leqslant 8 \\. \\ . \qquad \big(\frac{1}{2}\big)^x \leqslant 8 \\. \\ . \qquad (2^{-1})^x \leqslant 2^3 \\. \\ . \qquad 2^{-x} \leqslant 2^3 math . math \\ . \qquad -x \leqslant 3 \qquad \text{dividing by a negative so swap direction of sign} \\. \\ . \qquad x \geqslant -3 \\. \\ . \qquad x \geqslant -3.000 \qquad \textit{correct to 3 decimal places} math

NOTE The screen display on the right shows what the calculator gives as the exact answer for part //**a**//. The calculator defaults to ln(x) which is log(base **//e)//** when finding exact answers for inidicial equations.

The solution will be correct using any base as long as we are consistent with both numerator and denominator. We will cover base **//e//** later in the chapter. (See 3E equations with e)

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