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Solving Exponential and Logarithmic Equations: Techniques and Examples

This guide provides strategies for solving exponential and logarithmic equations effectively. It covers the properties of exponents and logarithms, emphasizing how to solve equations by equating exponents or taking logarithms when necessary. It includes multiple examples for clarity, such as solving 4^(3x) = 8^(x+1) and using logarithmic properties like log_b(x) = log_b(y) to find solutions. The importance of checking solutions for extraneous results is also highlighted. This resource is ideal for students seeking to strengthen their understanding of exponential and logarithmic functions.

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Solving Exponential and Logarithmic Equations: Techniques and Examples

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  1. 3.4Solving Exponential and Logarithmic Equations

  2. Exponential Equations • One way to solve exponential equations is to use the property that if 2 powers w/ the same base are equal, then their exponents are equal. • For b>0 & b≠1 if bx = by, then x=y

  3. Solve by equating exponents • 43x = 8x+1 • (22)3x = (23)x+1 rewrite w/ same base • 26x = 23x+3 • 6x = 3x+3 • x = 1 Check → 43*1 = 81+1 64 = 64

  4. Your turn! • 24x = 32x-1 • 24x = (25)x-1 • 4x = 5x-5 • 5 = x Be sure to check your answer!!!

  5. When you can’t rewrite using the same base, you can solve by taking a log or ln of both sides • 2x = 7 • log22x = log27 • x = log27 • x = ≈ 2.807

  6. 4x = 15 • log44x = log415 • x = log415 = log15/log4 • ≈ 1.95 Using Log Using LN

  7. 102x-3+4 = 21 • -4 -4 • 102x-3 = 17 • log10102x-3 = log1017 • 2x-3 = log 17 • 2x = 3 + log17 • x = ½(3 + log17) • ≈ 2.115

  8. Solving an Exponential Equation containing e • 40e0.6x-3= 237

  9. Solving Log Equations • To solve use the property for logs w/ the same base use the one-to one property: • If logbx = logby, then x = y

  10. log3(5x-1) = log3(x+7) • 5x – 1 = x + 7 • 5x = x + 8 • 4x = 8 • x = 2 and check • log3(5*2-1) = log3(2+7) • log39 = log39

  11. log5(3x + 1) = 2 Write in exponential form then solve! • 3x+1 = 25 x = 8 and check • Because the domain of log functions doesn’t include all reals, you should check for extraneous solutions

  12. log5x + log(x+1)=2 • log (5x)(x+1) = 2 (product property) • log (5x2 – 5x) = 2 • 5x2 - 5x = 100 • x2 – x - 20 = 0 (subtract 100 and divide by 5) • (x-5)(x+4) = 0 x=5, x=-4 • graph and you’ll see 5=x is the only solution

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