logs

[xc630]

Hey I need some help with this log problem

log4 (x+3) + log4 (x-3) = 2

WOuld the nexts ep be log 4 (x^2-9) = 2 ???

 

[stapel]

Assuming you mean the 4's to be the bases of the logs, the first step would be to apply the addition log rule to combine the terms on the left-hand side. After multiplying, you would obtain the equation you posted. The next step would be to convert the log equation to the equivalent exponential equation, and then solve the resulting quadratic equation for the value(s) of x.

Eliz.

. . . . ."log_b(x) = y" means the same as "b^y = x"

 

[mathmike]

log4 (x+3) + log4 (x-3) = 2

Do as Eliz advised except one word of caution: check that your solutions belong to the log domain. They must satisfy at the end both conditions:
x+3>0 and x-3>0 which in turn give one: x>3. Log of ... let's say (-8) doesn't exist for base 4.

 

[xc630]

I am confused. I don't know how to convert the log into an equivalent exponential equation. So I would change the left side of the equation
log 4 (x^2-9) = 2?

 

[tsh44]

I got x= 5

 

[mathmike]

I am confused. I don't know how to convert the log into an equivalent exponential equation.

Just remember that log is the POWER to which we have to raise base to get the argument. Consider:
log2(x) = 4
Using definition you can say that you have to raise 2 in power of 4 in order to get x. So, we can write:
x = 2^4
x = 16

some more examples:

log2(1/2) = -1
log2(sqrt(2)) = 1/2

The definition also explains domain of the log - there's no power in which you can raise 2 to get number <=0 as a result. So, following thing doesn't exist:
log2(-4).

 

[mathmike]

I got x= 5

Congratulations! Show your work, though for the benefit of other readers.