Basic Logarithms

Basic Logarithms

Solving equations involving exponents and logarithms Lets review some terms. When we write log5 125 5 is called the base 125 is called the argument

2 Logarithmic form of 5 = 25 is log525 = 2 For all the laws a, M and N > 0 a1 r is any real

Remember ln and log ln is a short cut for loge log means log10

Log laws log a 1 0 log a a 1 r log a M r log a M log a MN log a M log a N M log a

N log a M log a N ln M log a M ln a If your variable is in an exponent or

in the argument of a logarithm Find the pattern your equation resembles e b n

M N log b n e ln M ln N If your variable is in an exponent or in the argument of a logarithm Find the pattern

e b n M N

log b n e ln M ln N 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)

x ln(5) 2 log(2x) = 3 x ln(5)

2 log(2x) = 3 It fits log b n e x ln(5) 2

log(2x) = 3 10 =2x 3 Did you remember that log(2x) means log10(2x)? e

Switch b n log b n e x ln(5) 2 log(2x) = 3

10 =2x 3 500 = x Divide by 2 ln(x+3) = ln(-7x) ln(x+3) = ln(-7x)

It fits ln M ln N ln(x+3) = ln(-7x) Switch M N ln M ln N

ln(x+3) = ln(-7x) x + 3 = -7x Switch M N ln M ln N ln(x+3) = ln(-7x) x + 3 = -7x

x=- Solve the result (and check) ln(x) + ln(3) = ln(12) ln(x) + ln(3) = ln(12) x + 3 = 12 ln(x) + ln(3) = ln(12)

x + 3 = 12 Oh NO!!! Thats 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 M N ln M ln N ln(x) + ln(3) = ln(12)

ln(3x) = ln (12) 3x = 12 x=4 Solve the result log3(x+2) + 4 = 9 log3(x+2) + 4 = 9

It will fit log b n e log3(x+2) + 4 = 9 log3(x+2) = 5 Subtract 4 to make it fit log b n e

log3(x+2) + 4 = 9 log3(x+2) = 5 Switch e log b n e b n log3(x+2) + 4 = 9 log3(x+2) = 5 35 = x + 2

Switch e log b n e b n log3(x+2) + 4 = 9 log3(x+2) = 5 35 = x + 2 x = 241

Solve the result 5(10x) = 19.45 5(10x) = 19.45 10x = 3.91 Divide by 5 toe fit b n

5(10x) = 19.45 10x = 3.91 Switch e b n log b n e 5(10x) = 19.45

10x = 3.91 log(3.91) = x Switch e b n log b n e 5(10x) = 19.45 10x = 3.91

log(3.91) = x 0.592 Exact log(3.91) Approx 0.592 2 log3(x) = 8

2 log3(x) = 8 It will fit log b n e 2 log3(x) = 8 log3(x) = 4 Divide by 2 to fit

log b n e 2 log3(x) = 8 log3(x) = 4 Switch e log b n e b n

2 log3(x) = 8 log3(x) = 4 34=x Switch e log b n e b n 2 log3(x) = 8 log3(x) = 4 34=x

x = 81 Then Simplify log2(x-1) + log2(x-1) = 3 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 log MN log M log N log2(x-1) + log2(x+1) = 3 log2{(x-1)(x+1)} = 3 Switch e

log b n e b n log2(x-1) + log2(x+1) = 3 log2{(x-1)(x+1)} = 3 23=(x-1)(x+1) Switch e log b n e b n

log2(x-1) + log2(x+1) = 3 log2{(x-1)(x+1)} = 3 23=(x-1)(x+1) = x2 -1 and finish x = +3 or -3 log2(x-1) + log2(x+1) = 3 log2{(x-1)(x+1)} = 3 23=(x-1)(x+1) = x2 -1

But -3 does x = +3 or -3 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) ln x 3 1 Theres

more than one way to do this ln x 3 1 1 2 ln( x 3) 1 1 ln( x 3) 1

2 ln( x 3) 2 e 2 x 3 2 e 3 x x 4.389 Can you find why

each step is valid? ln x 3 1 1 2 ln( x 3) 1 1 ln( x 3) 1

2 ln( x 3) 2 2 e x 3 rules of exponents r log a M r log a M

multiply both sides by 2 e b n log b n e e 3 x - 3 to get exact answer

x 4.389 Approximate answer 2 ln x 3 1 1 e x 3

2 e x 3 2 e x 3 2

e 3 x x 4.389 2 Heres another way to

solve the same equation. ln x 3 1 1 e x 3 2

e x 3 2 Square both sides

2 e x 3 2 Simplify 2 e x 3 or e x 3 2

2 e 3 x or e 3 x 2 x e 3 4.389 exclude 2nd result

52x - 5x 12 = 0 52x - 5x 12 = 0 (5x 4)(5x + 3) = 0 Factor it. Think of 2 y - y-12=0

52x - 5x 12 = 0 (5x 4)(5x + 3) = 0 5 4 = 0 or 5 + 3 = 0 x x Set each factor = 0

Solve 5x 4 = 0 5 =4 x log54 = x Solve first factors equation Solve 5x + 3 = 0

5 = -3 x log5(-3) = x Solve other factors equation Solve 5x + 3 = 0 5 = -3

x log5(-3) = x Oops, we cannot have a negative argument

Solve 5x + 3 = 0 5 = -3 x log5(-3) = x Exclude this solution. Only the other factors solution

works x ln(5) 2 4x+2 = 5x x

ln(5) 2 4x+2 = 5x If M = N then ln M = ln N x

ln(5) 2 4 =5 x+2 x ln(4 ) = ln(5 ) x+2 x

If M = then N ln M = ln N 4 =5 x+2

x ln(4 ) = ln(5 ) x+2 x log a M r r log a M 4x+2 = 5x ln(4x+2) = ln(5x)

(x+2)ln(4) =(x) ln(5 ) log a M r r log a M 4x+2 = 5x ln(4x+2) = ln(5x) (x+2)ln(4) =(x) ln(5 ) x ln (4) + 2 ln(4) = x ln (5) Distribute

4x+2 = 5x ln(4x+2) = ln(5x) (x+2)ln(4) =(x) ln(5 ) x ln (4) + 2 ln(4) = x ln (5) x ln (4) - x ln (5)= 2 ln(4) Get x terms on one side

4x+2 = 5x ln(4x+2) = ln(5x) (x+2)ln(4) =(x) ln(5 ) x ln (4) + 2 ln(4) = x ln (5) x ln (4) - x ln (5)= 2 ln(4) x [ln (4) - ln (5)]= 2 ln(4) Factor out x

4x+2 = 5x ln(4x+2) = ln(5x) (x+2)ln(4) =(x) ln(5 ) x ln (4) + 2 ln(4) = x ln (5) x ln (4) - x ln (5)= 2 ln(4) x [ln (4) - ln (5)]= 2 ln(4) 2 ln(4) x ln(4) ln(5)

Divide by numerical coefficient

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