04 5indexeqns

Exponential Equations

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

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

Other exponential equations can be solved by taking the log of both sides, then simplifying.

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Example 1

Solve the following:

a)$3^{x-1} = 81$

b)$4^{2x-1} = \dfrac{1}{256}$

Solution:

a)$3^{x-1} = 81$

… … … $3^{x-1}=3^4$ … … equate the indices

… … … $x-1=4$

… … … $x=5$
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b)$4^{2x-1} = \dfrac{1}{256}$

… … … $4^{2x-1}=\dfrac{1}{4^4}$

… … … $4^{2x-1}=4^{-4}$ … … equate the indices

… … … $2x-1=-4$

… … … $2x=-3$

… … … $x=-\dfrac{3}{2}$

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Example 2

Find x, where: $3^{2x}-3^{x+2}+8=0$

Solution:

… … $3^{2x}-3^{x+2}+8=0$

… … use index laws to isolate $3^x$

… … $\big( 3^x \big)^2 - 3^x \times 3^2 + 8 = 0$

… … Let $u = 3^x$

… … $u^2-9u+8=0$ … … this is now a quadratic so factorise and solve

… … $(u-1)(u-8)=0$

… … $u=1 \;\; \text{ or } \;\; u=8$

… … substitute $u = 3^x$

… … $3^x=1 \;\; \text{ or } \;\; 3^x=8$

… … $x=0 \;\;\text{ or } \;\; x= \log_3(8)$ … … exact answer

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More Exponential Equations

Sometimes exponential equations can be solved by taking the log of both sides.
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Example 3

Solve $32^x = 28$

a) … in exact form using base 10

b) … correct to 3 decimal places

Solution:

… … $3^{2x}=28$

… … Take the log(base 10) of both sides

… … $\log_{10}\big( 3^{2x} \big) = \log_{10}(28)$

… … $2x \log_{10}(3) = \log_{10}(28)$

… … $2x =\dfrac{\log_{10}(28)}{\log_{10}(3)}$

… … $x = \dfrac{\log_{10}(28)}{2 \log_{10}(3)}$ … … exact answer

… … $x = 1.517$ … … … correct to 3 decimal places

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  • This can be calculated using the ClassPad, either from the final answer in part (a),
  • or by solving the initial equation.

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Example 4

Solve the following, answering in both exact form and correct to 3 decimal places:

a)$5^{3x} > 29$

b)$0.5^x \leqslant 8$

Solution:

a)$5^{3x} > 29$

… … … Take log (base 10) of both sides {or use log(base e), see below}

… … … $\log_{10} \big( 5^{3x}\big) > \log_{10}(29)$

… … … $3x \log_{10}(5) > \log_{10}(29)$

… … … $3x > \dfrac{ \log_{10}(29)}{\log_{10}(5)}$

… … … $x > \dfrac{\log_{10}(29)}{3\log_{10}(5)}$ … … exact answer

… … … $x > 0.697$ … … … correct to 3 decimal places

04.5eg4a.gif

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b)$0.5^x \leqslant 8$

… … … $\big(\frac{1}{2}\big)^x \leqslant 8$

… … … $(2^{-1})^x \leqslant 2^3$

… … … $2^{-x} \leqslant 2^3$ … … equate the exponents

… … … $-x \leqslant 3$ … … divide by negative so swap direction of sign

… … … $x \geqslant -3$ … … exact answer

… … … $x \geqslant -3.000$ … … correct to 3 decimal places
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04.5eg4b.gif

NOTE: If you attempt to solve indicial equations on the Classpad, it defaults to ln(x) {which is log(base e)}.

  • 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.

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