Write as a single logarithm. log 2 + log x + log y Precalculus Section 5.6 Apply laws of logarithms http://www.youtube.com/watch?v=SxF44olWTyk Laws of exponents bx ∙by = bx+y bx/by = bx-y (bx)y = bxy If bx = by then x = y Laws of logarithms logb (xy) = logbx + logby logb (x/y) = logbx – logby logb (yx ) = x logb y If logbx = logby then x = y Write as a single logarithm. log 2 + log x + log y Answer: log(2xy)

Write as a single logarithm log 15 – log 5 Answer: log (15/5) log (3) 2 ln x + 3 ln y – 4 ln z Answer: ln(x2y3 / z4) log248 – 1/3 log2 27 Answer: log2(48/271/3) log2 (48/3) log2 (16) 4

Expand, write in terms of log x and log y log Answer: log x - 2 log y log Answer: ½ log x + 3/2 log y

Simplify ln Answer: 2x answer: 101 102 log x 10 x2 Answer: e2 ∙ e-ln x 2 x-1

Solve for y log y = 2 log x Answer: log y = log x2 y = x2 log y + ½ log x = log 3 Answer: log(yx1/2) = log 3 yx1/2 = 3 y = 3/ x1/2

assignment Page 200 Problems 3 – 30 ÷ 3, 31,40, 43-46