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 b > 0 and b cannot equal 1 (why not?)

Review with a POD What we’ve seen so far: If y = b x, then x = log b y. 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 = 8.5 t.

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

Solving for the exponent 1. log 2 8 = x 2. log 3 81 = x 3. log 4 32 = x

Solving for the exponent 1. log 2 8 = x 2 x = 8x = 3 using guess and check or common base You could also set it up with the change of base. 2. log 3 81 = x 3 x = 81x = 4ditto 3. log 4 32 = x 4 x = 32x = 2.5 ditto

Solving for the argument What is the argument again? 1. log 3 x = log 5 x = 5 3. log 4 x = 0 How could you check your answers?

Solving for the argument 1. log 3 x = = xx = 1/81 2. log 5 x = = xx = log 4 x = = xx = 1

Solving for the base What is the base again? 1. log x 8 = 3 2. log x 25 = 2/3 How could you check these answers?

Solving for the base 1. log x 8 = 3 x 3 = 8 (x 3 ) 1/3 = 8 1/3 x = 2 2. log x 25 = 2/3 x 2/3 = 25 (x 2/3 ) 3/2 = 25 3/2 x = 125