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Logarithms with Other Bases (6.9)

Logarithms with Other Bases (6.9). Solving the three parts of logarithmic equations. Review with a POD. What we’ve seen so far: If y = b x , then x = log b y. Vocabulary review: x is the b is the y is the There are certain conditions b and y must meet: y > 0

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Logarithms with Other Bases (6.9)

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  1. Logarithms with Other Bases (6.9) Solving the three parts of logarithmic equations

  2. Review with a POD What we’ve seen so far: If y = bx, then x = logby. Vocabulary review: x is the b is the y is the There are certain conditions b and y must meet: y > 0 b > 0 and b cannot equal 1 (why not?)

  3. Review with a POD What we’ve seen so far: If y = bx, then x = logby. When we write with logs we’re solving for the exponent: The exponent is by itself. b is the base (in the basement). Rewrite to solve for t: m = 8.5t.

  4. Review with a POD What we’ve seen so far: If y = bx, then x = logby. Rewrite these statements using logs: 10x = 5. 6x = 4/3 2x = 8 How would you solve any of them?

  5. Solving for the exponent • log2 8 = x • log3 81 = x • log4 32 = x

  6. Solving for the exponent • log2 8 = x 2x = 8 x = 3 using guess and check or common base You could also set it up with the change of base. • log3 81 = x 3x = 81 x = 4 ditto • log4 32 = x 4x = 32 x = 2.5 ditto

  7. Solving for the argument What is the argument again? • log3 x = -4 • log5 x = 5 • log4 x = 0 How could you check your answers?

  8. Solving for the argument • log3 x = -4 3-4 = x x = 1/81 • log5 x = 5 55 = x x = 3125 • log4 x = 0 40 = x x = 1

  9. Solving for the base What is the base again? • logx 8 = 3 • logx 25 = 2/3 How could you check these answers?

  10. Solving for the base • logx 8 = 3 x 3 = 8 (x3)1/3 = 81/3 x = 2 • logx 25 = 2/3 x 2/3 = 25 (x2/3)3/2 = 253/2 x = 125

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