Logs: Solve log_2(x) + log_6(x + 1) = log_6(1) for x

[Gadsilla]

Solve for x:

log₂x+log₆(x+1)=log₆1


So I've been doing a lot of log exercises, they are mostly like these:

log₃x-log₃(5)=4

and

log₂(3x+2)=2

All solving for X. The exampled ones are all fine, however I am unsure how to go forward with the one in question.

So far I've done this

log₂x+log₆(x+1)=log₆1

log₂x(x(x+1)) = log₆1


Here's where I get stuck if I'm doing it right ?

 

[Harry_the_cat]

Solve for x:

log₂x+log₆(x+1)=log₆1


So I've been doing a lot of log exercises, they are mostly like these:

log₃x-log₃(5)=4

and

log₂(3x+2)=2

All solving for X. The exampled ones are all fine, however I am unsure how to go forward with the one in question.

So far I've done this

log₂x+log₆(x+1)=log₆1

log₂x(x(x+1)) = log₆1


Here's where I get stuck if I'm doing it right ?

Are you sure you've typed the original question correctly? with log base 2 and log base 6?

 

[pka]

Solve for x:
{tex]\log_2(x)+\log_6(x+1)=\log_6(1)[/tex]
So I've been doing a lot of log exercises, they are mostly like these:

If THAT is true then you should have learned that:
if \(\displaystyle b>0\) then \(\displaystyle \log_b(1)=0\) , moreover you should have learn that your problem is equivalent to:

\(\displaystyle \dfrac{\log(x)}{\log(2)}+\dfrac{\log(x+1)}{log(6)}=0\)

 

[Gadsilla]

If THAT is true then you should have learned that:
if \(\displaystyle b>0\) then \(\displaystyle \log_b(1)=0\) , moreover you should have learn that your problem is equivalent to:

\(\displaystyle \dfrac{\log(x)}{\log(2)}+\dfrac{\log(x+1)}{log(6)}=0\)

Thanks.

 

[Steven G]

Thanks.

So you did solve this in the end? What was the common denominator you used? Can you show us your work? This is a difficult problem and we just want to make sure you did it correctly. If you in fact need more help on this we will help you. That is never a problem.

 

[Gadsilla]

So you did solve this in the end? What was the common denominator you used? Can you show us your work? This is a difficult problem and we just want to make sure you did it correctly. If you in fact need more help on this we will help you. That is never a problem.

Yes, I'm mostly fine with logs.