Thread: I'm having trouble with these logs...not sure how to get started.

1. I'm having trouble with these logs...not sure how to get started.

log4^(x+7)-log4^(x-3)

I may have figured out this one...

(x+7)-(x-3)
x=-(-7-3)
x=10
Is this right???

and

logc^x+logc^y

also this one, but it's different from the first two i think...any help getting started would be terrific!

log2^(x+1)+log2^(x-5)=4

I may have figured out the last problem...

log2^(x+1)+log2^(x-5)=4
log2x+log2(1)=log2+x+log2-5=4
x+1+x-5=4
x-4=4
x=0
??? Is this right ???

2. Log4(x+7) - log4(x-3) = Log4((x+7)/(x-3)), i don't know how far you have to go with this problem, however you can simplify it to

Log4( 1 + 10/(x+3))

Logc(x) + logc(y) = Logc(xy), again im not quite sure what they want you to do with this. If it is Logc(xy) = 0 then it can be simplified to xy = 1... so y = 1/x??

Log2(x+1) + log2(x-5) = 4;
Log2((x+1)(x-5)) = 4;
Log2(x^2 - 4x - 5) = 4;
x^2 - 4x - 5 = 2^4;
x^2 - 4x - 5 = 16;
x^2 - 4x - 21 = 0;
(x-7)(x+3) = 0, hence x = 7 or -3

Good luck

3. Originally Posted by Mhockey02
log4^(x+7)-log4^(x-3)

I may have figured out this one...

(x+7)-(x-3)
x=-(-7-3)
x=10
Is this right???,
You must have an equation to end up with an equation. Since you don't know what the original expression equals, so you can't solve it. The expression, however, can be simplified.

But all these probelems are asking about properties of logs. With logs, there are three properties you should know (in addition to the change base formula). These are:
$\displaystyle log_b(a^x) = x log_b(a)$
$\displaystyle log_b(m) + log_b(n) = log_b(mn)$
$\displaystyle log_b(m) - log_b(n) = log_b(\frac{m}{n})$

$\displaystyle log[4^{x+7}]-log[4^{x-3}]$ can become $\displaystyle log[\frac{4^{x+7}}{4^{x-3}}]$. Of course, when you are dividing with a common base, you can subtract exponents. Doing that would give you $\displaystyle log[4^{(x+7)-(x-3)}$ or $\displaystyle log[4^{10}]$.

Plugging this into a calculator, I get $\displaystyle log[4^{x+7}]-log[4^{x-3}]=6.0205999$ (approximately).

Originally Posted by Mhockey02
logc^x+logc^y
You can use similar logic for this one. Another way to do both of these problems is to apply the first property (the 'exponents' property) first:

$\displaystyle log[c^x]+log[c^y]=xlog(c)+ylog(c)=(x+y) log(c)$

Originally Posted by Mhockey02
also this one, but it's different from the first two i think...any help getting started would be terrific!

log2^(x+1)+log2^(x-5)=4

I may have figured out the last problem...

log2^(x+1)+log2^(x-5)=4
log2x+log2(1)=log2+x+log2-5=4
x+1+x-5=4
x-4=4
x=0
??? Is this right ???
Nope, not correct.

$\displaystyle log[2^{x+1}]+log[2^{x-5}]=4$ (given)
$\displaystyle (x+1)log(2)+(x-5)log(2)=4$ ('exponents' property)
$\displaystyle x+1+x-5=\frac{4}{log(2)}$ (dividing by $\displaystyle log(2)$)
$\displaystyle 2x-4=\frac{4}{log(2)}$ (simplifying)
$\displaystyle 2x=\frac{4}{log(2)}+4$ (adding 4)
$\displaystyle x=\frac{2}{log(2)}+2$ (dividing by 2)
This is approximately x=8.644

4. Originally Posted by spacemonkey
Log2(x+1) + log2(x-5) = 4;
Log2((x+1)(x-5)) = 4;
Log2(x^2 - 4x - 5) = 4;
x^2 - 4x - 5 = 2^4;
x^2 - 4x - 5 = 16;
x^2 - 4x - 21 = 0;
(x-7)(x+3) = 0, hence x = 7 or -3
If this IS what is meant by the problem (log base 2), then this is a good method. Be careful, however, because you CAN get extraneous solutions. In this case, the -3 is extraneous. You cannot have a log of 0 or the log of a negative. If x=-3, then you've got the log of -2 and the log of -8 in the original equation. This doesn't work. The only solution that works is 7.

5. Originally Posted by pflo
You cannot have a log of 0 or the log of a negative. If x=-3, then you've got the log of -2 and the log of -8 in the original equation. This doesn't work. The only solution that works is 7.
yes you are absolutely right