finding real roots of a log equation ... not sure what to do

[mathstudent1998]

Explain why the equation


2log(x+3)+logx−log(4x+2)=1

(all the logs are in base2)


has only one real root and state the exact value of this root.



I'm an A level math and physics student, I just completed my first year of A level, I studied c1, c2 and s1 of ccea AS modules (if that helps) ...

Im attending a summer school on thursday and the university have sent me some questions to have a go at, im finding most of them very easy, but im not sure how to start this particular question.

 

[Ishuda]

Explain why the equation


2log(x+3)+logx−log(4x+2)=1

(all the logs are in base2)


has only one real root and state the exact value of this root.

What are your thoughts? What have you done so far? Please show us your work even if you feel that it is wrong so we may try to help you. You might also read
http://www.freemathhelp.com/forum/threads/78006-Read-Before-Posting


As a hint, remember the rules for logs such as
a log(x) = log(x^a)
log(x) + log(y) = log (x * y)
and
log_a(x) = y -----> x = a^y

EDIT: Oh, and because of the way the problem is stated I strongly suspect they mean "has only one real root in the domain of the base 2 log function".

 

[stapel]

Explain why the equation

\(\displaystyle 2\,\log_2(x\,+\,3)\,+\,\log_2(x)\,−\,\)\(\displaystyle \log_2(4x\,+\, 2)\,=\,1\)

has only one real root and state the exact value of this root.

What are your thoughts? What have you tried? How far have you gotten? Where are you stuck?

For instance, you first applied log rules (here) to combine the terms on the left-hand side. Then you applied the basic definitional relationship between logs and exponentials (here) to convert the equation into exponential form. You then simplified to get a rational equation, which you solved in the usual manner (here). (You probably did a graph to find useful test values for possible solutions of the cubic polynomial, or used other methods (here), possibly numerical ones (here).) Then you checked your solution(s) against the domains of the logs in the original equation, and... then what?

Please be complete. Thank you!

 

[mathstudent1998]

actually i'm mainly stuck with the expansion of the brackets since they're logarithms and then not sure how to put it into an equation with only 3 parts so i can use the b^2-4ac >0 to get the real root

here's what i've attempted so far ..

I brought the 1 over and made the equation equal to 0...
2\log2(x+3) +log2(x) - log2(4x+2) -1 = 0

im not sure how to expand the brackets of the logs .. but i ended up with
2\log2(x) + 2\log2(3) + log2(x) - log2(4x) + log2(2) -1 =0

the collected like terms ...
3\log2(x) + 2\log2(3x2=6) -log2(4x) -1 =0

but i think i've already made a mistake at that point

 

[mathstudent1998]

sorry i replied a while ago, but it wont post..

Anyway, Im mainly stuck on how to expand the brackets and putting the equation into one where i can use the b^2-4ac>0 to find the real root.

so far i've brought the 1 over from the left hand side and put it on the right so the equation is
2\log2(x+3) + ... -1 =0

i've treid expanding the brackets a few times but i'm still not sure where to start with it

 

[mathstudent1998]

What are your thoughts? What have you tried? How far have you gotten? Where are you stuck?

For instance, you first applied log rules (here) to combine the terms on the left-hand side. Then you applied the basic definitional relationship between logs and exponentials (here) to convert the equation into exponential form. You then simplified to get a rational equation, which you solved in the usual manner (here). (You probably did a graph to find useful test values for possible solutions of the cubic polynomial, or used other methods (here), possibly numerical ones (here).) Then you checked your solution(s) against the domains of the logs in the original equation, and... then what?

Please be complete. Thank you!

I figured it out now, but thanks for the help