Algebra 2

posted by .

log4 (x-2) - log4 (x+1) = 1

  • Algebra 2 -

    log4(x-2)(x+1) = 1
    (x-2)(x+1) = 4
    x^2 - x - 6 = 0
    (x-3)(x+2)
    x = 3 or -2
    however, x = -2 does not fit the original equation, since log of negative numbers are not defined.

    So, the only solution is x=3

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Algebra II

    Solve, round to four decimal places if necessary: log4(x-9)=2 Answer: x-9=4^2 x-9=16 x=11 log4(z)+log4(z-3)=1 Answer: z+z-3=1 2z=4 z=2 Thanks for checking my work.
  2. Logarithms

    Solve log4 x + log4(x-2)=log4(15) I know how do the example in my book but I don't know what I'm doing wrong here. log4 x + log4(x-2)=log4(15) log4 [x + log4(x-2)]=log4(15) (x^2 + x)=4 x^2 + x-4=0
  3. math

    Solve the equation algebraically. Round the result to three decimal places. log4 x - log4(x-1) = 1/2 I have no idea how to do this. Please help!
  4. Algebra 2

    Simplify: 1. 2log10^5+log10^4 2. 2log3^6-log3^4 3. log4^40-log4^5 4. log4^3-log4^48 I need to see all of the steps. Thanks
  5. Math

    Evaluate the expression without using a calculator: log4(2)= ____?
  6. Math(Please check)

    solve for x: log4 X - log4 2 = 2 log(2x) = 2 Now what do I do?
  7. Math

    solve for x: log4 X - log4 2 = 2 log(2x) = 2 Now what do I do?
  8. Logarithm help

    Write each expression as a single logarithm I need help really fast. I have to finish this by today. log4^3 + 1/2log4^8 - log4^2
  9. precalc

    I don't understand how to do these w/o calc. I tried to write it in a way that will make someone understand how to read it. Hope I typed it clear enough.Thanks so much for the help anyone! How to find the exact value of logarithm: …
  10. Precalc

    Using logarithms to rewrite the expression: 1.) log4.(4)^3x= 2.) 6 log6 36= 3.) 3 log 2 1/2= 4.) 1/4 log4 16

More Similar Questions