Post a New Question

Algebra 2

posted by on .

Can someone please explain to me how to solve logarithmic equations? Some examples are: -5=log2X
3^5x+1= 40 and log6X+(log6)7=(log6)35
I hope that makes sense because logs are hard to type.

  • Algebra 2 - ,

    -5=log2X

    -5 = log2x

    2^-5 = x
    x = 1/32


    3^(5x+1)= 40
    let's take the log of both sides

    log(3^(5x+1)) = log40
    (5x + 1) log3 = log40
    5x+1 = log40/log3
    etc.

    log6X+(log6)7=(log6)35
    log6(7x) = log6 35
    "antilog" both sides
    7x = 35
    etc.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question