1. ## Logarithms problems

hey i just have no clue in how to attempt this,
anyone shine a light?
Q: solve equations
a)logx+3 = 4
b) log(x+3) = 4
C) logx+log3 = 4

Q: Log(x+1) - log(x-6) +2logx i know here you have apply the rules of log, but i have not really been used to dealing with brackets in these situations

hey i just have no clue in how to attempt this,
anyone shine a light?
Q: solve equations
a)logx+3 = 4
b) log(x+3) = 4
C) logx+log3 = 4

Q: Log(x+1) - log(x-6) +2logx i know here you have apply the rules of log, but i have not really been used to dealing with brackets in these situations

Looks like you need a text book or a good tutorial. Try this one: College Algebra Tutorial on Logarithmic Equations

I'll do the first one, and get you started on the other two.

a) $\displaystyle \log x+3=4$

$\displaystyle \log x = 1$

$\displaystyle x=10^1=10$

b) $\displaystyle \log (x+3)=4$

$\displaystyle x+3=10^4$

c) $\displaystyle \log x+\log 3 =4$

$\displaystyle \log 3x=4$

$\displaystyle 3x=10^4$

3. thankyou=] ill have a look at the help guide.
much appreciated

4. hey with this equation what would i do first
or does it not matter if i work with the product rule then the quotient rule or vis versa, i=or is one way easier.
as with the examples on the website you showed me, when they were working there equations out, they seemed to be left with a exponent on the other side of the equals sign, and unforunatly i have not got that on this equation
log(x+1) - log(x-6) +2logx
at first i did
log (x+1/x-6) +log x^2
and im not sure but is there a way to make the power (2) to become the exponent or am i talking nonsence, as if i could i would do
somehow
log(X+1/X-6)=2
then i could
10^2 = x+1/x-6
and then from that solve x
but im not sure
?

hey with this equation what would i do first
or does it not matter if i work with the product rule then the quotient rule or vis versa, i=or is one way easier.
as with the examples on the website you showed me, when they were working there equations out, they seemed to be left with a exponent on the other side of the equals sign, and unforunatly i have not got that on this equation
log(x+1) - log(x-6) +2logx
at first i did
log (x+1/x-6) +log x^2
and im not sure but is there a way to make the power (2) to become the exponent or am i talking nonsence, as if i could i would do
somehow
log(X+1/X-6)=2
then i could
10^2 = x+1/x-6
and then from that solve x
but im not sure
?
First of all, $\displaystyle \log(x+1) - \log (x-6) +2 \log x$ is not an equation.

Did you miscopy it? There's no '=' sign.

6. =] that may be why what i was doing wasn't making alot of sence.
so would it still be
log(X+1/X-6) +logx^2
then
logx^3+x^2 divided by x-6
?