College Algebra
 👍 0
 👎 0
 👁 133

 👍 0
 👎 0
Respond to this Question
Similar Questions

Math
If ln a=2, ln b=3, and ln c=5, evaluate the following: (a) ln(a^−2/b^4c^−3)= .9524 (b) ln(√b^−4*c^1*a^−1)= 1.739 (c) ln(a^2b^4)/ln(bc)^−2= 42.39 (d) (ln(c^−1))*(ln(a/b^3))^−4= .03507 I am getting these answers

Algebra2
Expand the logarithmic expression. log8 a/2 My answer is log8alog8 2

Algebra 2
How is the expression log 32  log 8 written as a single logarithm? log 4 log 8 log 24 log 40

Math
Log8 4 + log16 2 x log 27 81= x solve for x

Chemistry
A galvanic cell is based on the following halfreactions at 25C: Ag+ + e > Ag H2O2 + 2H+ + 2e > 2H2O (not H2O2) Predict whether E_cell is larger or smaller than E°_cell for the following case. [Ag+]= 1.0M, [H2O2]= 2.0 M,

Trigonometry
This is a logs question If u=x/y^2, which expression is equivalent to log u? 1) log x + 2 log y 2) 2(log x log y) 3) 2(log x + log y) 4) log x 2 log y

Math
if log(x+1) 27=3 find the value of x *(x+1) is the base i know that the answer is 2 but i have no clue how to get that

Pre Calculus
Evaluate log b {square root of 10b}, given that log b 2 = 0.3562 and log b 5 = 0.8271 (sorry I had no clue how to type this in math format!)

math
how do i solve for x when x = log(32.5)? here's my work: log x = y 10^y = x then, x = log (32.5) 10^x = 32.5 10^x = 17/2 this is where i get stuck... i don't know how to get rid of the bases... can some one please teach me how?

Trigonometry
The expression log(x^n/ radical y)is equivalent to 1. n log x  1/2 log y 2. n log x 2 log y 3. log (nx)  log (1/2y) 4. log (nx)  log (2y)

Pre Calculus
Assume that x, y, and b are positive numbers. Use the properties of logarithms to write the expression logb ^8xy in terms of the logarithms of x and y. a. logb^8 + logb x + logb^y b. logb^8+logbx c. logb^8+logby d. logb^8 + log8 x

math(Please help)
1) use the properties of logarithms to simplify the logarithmic expression. log base 10 (9/300) log  log 300 log 9 = 2 log 3 log 300 = log 3 + log 100 = log 3+2 I just do not know how to put these together now!
You can view more similar questions or ask a new question.