College Algebra & Trigonometry, 2018a

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3.3. SOLVING EXPONENTIAL EQUATIONS 169 If we collect like terms, we’ll end up with: x ∗ 3log4=x ∗ 2log9− log 9 log 9 = x ∗ 2log9− x ∗ 3log4 At this point, if we want to get the x by itself, we need to factor out the x on the right-hand side: log 9 = x ∗ 2log9− x ∗ 3log4 log 9 = x(2 log 9 − 3 log 4) Then divide on both sides by the coefficient in parentheses: log 9 2log9− 3log4 = x ✭✭✭✭✭✭✭✭✭ (2 log 9 − 3 log 4) ✭✭✭✭✭✭✭✭ 2log9− 3log4 log 9 2log9− 3log4 = x 9.327 ≈ x Again, we can check our answer by plugging it back into the equation: 4 3∗9.327 ≈ 4 27.981 ≈ 7.0184 ∗ 10 16 9 2∗9.327−1 ≈ 9 17.654 ≈ 7.0177 ∗ 10 16
170 CHAPTER 3. EXPONENTS AND LOGARITHMS We could also have solved this equation by approximating the logarithms in the beginning. 4 3x =9 2x−1 log 4 3x =log9 2x−1 3x ∗ log4=(2x − 1) log 9 3x(0.60206) ≈ (2x − 1)0.95424 1.80618x ≈ 1.9085x − 0.95424 0.95424 ≈ 0.10232x 9.326 ≈ x This answer is less accurate than the other approximation (9.326036 vs. 9.327424). The accuracy of an answer depends upon the original approximations for the logarithms.
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College Algebra & Trigonometry
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c○2018 Richard W. Beveridge This
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4 CONTENTS 2.4 Solution of Rational
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6 CONTENTS 8.2 The Trigonometric Ra
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College Algebra & Trigonometry, 2018a

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