Log Help: log6(b2+2)+log62 = 2, log3(5x+5)-log3(x2-1) = 0,  TOPIC_SOLVED

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.

Log Help: log6(b2+2)+log62 = 2, log3(5x+5)-log3(x2-1) = 0,

Postby klongoria1 on Tue Apr 27, 2010 1:24 pm

What would it look like to solve this equation, steph by step? This problem was in a math lesson of mine in which I was taugth the Properties of Logarithms: the Product, Quotient, and Power properties.

log6(b2 + 2) + log62 = 2

And this one, the zero at the end is really throwing me off:

log3(5x+5) - log3(x2 - 1) = 0

Also, how do you solve these 2 problems, and what is the main difference?

Equation 1: log 2 + log x = log 3

Equation 2: log 2 + log x = 3

Thanks!!
klongoria1
 
Posts: 5
Joined: Fri Apr 16, 2010 8:37 pm

Sponsor

Sponsor
 

Postby stapel_eliz on Tue Apr 27, 2010 4:06 pm

Do you know what logs are, and what the log rules are?
User avatar
stapel_eliz
 
Posts: 1784
Joined: Mon Dec 08, 2008 4:22 pm

Re: Log Help

Postby klongoria1 on Thu Apr 29, 2010 8:19 pm

I do, but I need help applying some of the rules.
klongoria1
 
Posts: 5
Joined: Fri Apr 16, 2010 8:37 pm

  TOPIC_SOLVED

Postby stapel_eliz on Fri Apr 30, 2010 12:23 pm

klongoria1 wrote:I do, but I need help applying some of the rules.

1) log6(b2 + 2) + log6(2) = 2

Use the "addition" log rule to combine the logs on the left-hand side. Then use The Relationship to convert the resulting log equation into the equivalent exponential form. Then solve the quadratic equation for the value(s) of b.

2) log3(5x + 5) - log3(x2 - 1) = 0

Move one of the logs to the other side of the "equals" sign, equate the arguments, and then solve the resulting quadratic equation.

3) log(2) + log(x) = log(3)

Use the same log rule you used in (1) to combine the logs on the left-hand side. Then equate and solve, just as you did in (2). (For loads of worked examples of solving log equations, try here.)

4) log(2) + log(x) = 3

Use the same log rule you used in (1) and (3) to combine the logs on the left-hand side. Then convert, as you did in (2), to the equivalent exponential form, and solve the resulting linear equation.

If you get stuck on any of the above, please reply showing how far you have gotten. Thank you! :wink:

Note: I'm afraid I don't know what you mean by "the main difference"...?
User avatar
stapel_eliz
 
Posts: 1784
Joined: Mon Dec 08, 2008 4:22 pm


Return to Advanced Algebra ("pre-calculus")