Download Exponential and Logarithmic Equations: Solved Examples and more Lecture notes Pre-Calculus in PDF only on Docsity!
Sample Exponential and Logarithm Problems
1 Exponential Problems
Example 1.1 Solve
)โ 3 xโ 2 = 36x+1.
Solution: Note that^16 = 6โ^1 and 36 = 6^2. Therefore the equation can be written
(6โ^1 )โ^3 xโ^2 = (6^2 )x+ Using the power of a power property of exponential functions, we can multiply the exponents: 63 x+2^ = 6^2 x+ But we know the exponential function 6x^ is one-to-one. Therefore the exponents are equal, 3 x + 2 = 2x + 2 Solving this for x gives x = 0.
Example 1.2 Solve 25 โ^2 x^ = 125x+7.
Solution: Note that 25 = 5^2 and 125 = 5^3. Therefore the equation is (5^2 )โ^2 x^ = (5^3 )x+ Using the power of a power property to multiply exponents gives 5 โ^4 x^ = 5^3 x+ Since the exponential function 5x^ is one-to-one, the exponents must be equal: โ 4 x = 3x + 21 Solving this for x gives x = โ.
Example 1.3 Solve exe^2 = (^) exe+1^4.
Solution: Using the product and quotient properties of exponents we can rewrite the equation as ex+2^ = e^4 โ(x+1) = e^4 โxโ^1 = e^3 โx Since the exponential function ex^ is one-to-one, we know the exponents are equal: x + 2 = 3 โ x
Solving for x gives x =^12.
2 Log Problems
Example 2.1 Wite the follwing equations in exponential form:
(a) 2 = log 3 9 (b) โ3 = loge e^13 (c) 12 = log 81 9 (d) log 4 16 = 2 (e) log 10 0 .0001 = โ 3
Solution: Use the correspondence loga y = x โโ y = ax:
(a) 2 = log 3 9 โโ 9 = 3^2 (b) โ3 = loge e^13 โโ (^) e^13 = eโ^3 (c) 12 = log 81 9 โโ 9 = 81^1 /^2 (d) log 4 16 = 2 โโ 16 = 4^2 (e) log 10 0 .001 = โ 3 โโ 0 .001 = 10โ^3
Example 2.4 Write the expression log 6 30 โ log 6 10 as a single term.
Solution: This just means use the quotient rule:
log 6 30 โ log 6 10 = log 63010 = log 6 3
Example 2.5 Solve log x โ 1 = โ log(x โ 9).
Solution: Put all logarigthms on the same side, and all numbers on the other side, so we can use the correspondence y = ax^ โโ loga y = x: log x + log(x โ 9) = 1 Use the product rule to simplify the left side, log(x(x โ 9)) = 1 Note, log y means the base is to be understood as 10, that is we have log 10 (x(x โ 9)) = 1 By the correspondence we know x(x โ 9) = 10^1 = 10 That is, x^2 โ 9 x โ 10 = 0 This polynomial factors: (x โ 10)(x + 1) = 0, so x = 10 or x = โ1. Looking at the original equation, we see we canโt use x = โ1, because log(โ1) and log(โ 1 โ 9) = log(โ10) are undefined. Thus, our only solution is x = 10
