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Learn how to solve equations involving exponents and logarithms with step-by-step explanations and examples. Understand logarithmic forms, log laws, and how to fit equations to patterns.
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Let’s review some terms.When we write log5 1255 is called the base125 is called the argument
For all the lawsa, M and N > 0 a≠ 1 r is any real
Remember ln and log • ln is a short cut for loge • log means log10
Find the pattern your equation resembles If your variable is in an exponent or in the argument of a logarithm
Find the pattern Fit your equation to match the pattern Switch to the equivalent form Solve the result Check (be sure you remember the domain of a log) If your variable is in an exponent or in the argument of a logarithm
log(2x) = 3 It fits
log(2x) = 3 103=2x Switch Did you remember that log(2x) means log10(2x)?
log(2x) = 3 103=2x 500 = x Divide by 2
ln(x+3) = ln(-7x) It fits
ln(x+3) = ln(-7x) Switch
ln(x+3) = ln(-7x) x + 3 = -7x Switch
ln(x+3) = ln(-7x) x + 3 = -7x x = - ⅜ Solve the result (and check)
ln(x) + ln(3) = ln(12) x + 3 = 12
ln(x) + ln(3) = ln(12) x + 3 = 12 Oh NO!!! That’s wrong!
ln(x) + ln(3) = ln(12) ln(3x) = ln (12) You need to use log laws
ln(x) + ln(3) = ln(12) ln(3x) = ln (12) 3x = 12 Switch
ln(x) + ln(3) = ln(12) ln(3x) = ln (12) 3x = 12 x = 4 Solve the result
log3(x+2) + 4 = 9 It will fit
log3(x+2) + 4 = 9 log3(x+2) = 5 Subtract 4 to make it fit
log3(x+2) + 4 = 9 log3(x+2) = 5 Switch
log3(x+2) + 4 = 9 log3(x+2) = 5 35 = x + 2 Switch
log3(x+2) + 4 = 9 log3(x+2) = 5 35 = x + 2 x = 241 Solve the result
5(10x) = 19.45 10x = 3.91 Divide by 5 to fit
5(10x) = 19.45 10x = 3.91 Switch
5(10x) = 19.45 10x = 3.91 log(3.91) = x Switch
5(10x) = 19.45 10x = 3.91 log(3.91) = x ≈ 0.592 Exact log(3.91) Approx 0.592
2 log3(x) = 8 It will fit
2 log3(x) = 8 log3(x) = 4 Divide by 2 to fit
2 log3(x) = 8 log3(x) = 4 Switch
2 log3(x) = 8 log3(x) = 4 34=x Switch
2 log3(x) = 8 log3(x) = 4 34=x x = 81 Then Simplify
log2(x-1) + log2(x-1)= 3 Need to use a log law
log2(x-1) + log2(x+1)= 3 log2{(x-1)(x+1)} = 3
log2(x-1) + log2(x+1)= 3 log2{(x-1)(x+1)} = 3 Switch
log2(x-1) + log2(x+1)= 3 log2{(x-1)(x+1)} = 3 23=(x-1)(x+1) Switch
log2(x-1) + log2(x+1)= 3 log2{(x-1)(x+1)} = 3 23=(x-1)(x+1) = x2 -1 x = +3 or -3 and finish
log2(x-1) + log2(x+1)= 3 log2{(x-1)(x+1)} = 3 23=(x-1)(x+1) = x2 -1 x = +3 or -3 But -3 does not check!
log2(x-1) + log2(x+1)= 3 log2{(x-1)(x+1)} = 3 23=(x-1)(x+1) = x2 -1 x = +3 or -3 Exclude -3 (it would cause you to have a negative argument)