logarithms problems,

Logarithms problems.

logarithms problems,

Postby Learningforcomp » Sat Sep 12, 2020 9:34 am

Let a and b two real numbers positives and different than 1. What is the relation between a and b so that the equation x^2-x(log(b)a)+2log(a)b=0 will have two equal roots
The answer is a^2=b
To solve it i used the delta expression of the baskhara formula: b^2-4*a*c since it needs to have the same the same two roots it means delta=0

So I applied the formula and got to log^2(a)b-8log(b)a=0 ==> [log(a)b]^2=8log(b)a

- taking the SQUARE root of both sides i got log(a)b=log(b)a
- I can apply change of base to log(b)a and the end result will be: log(a)b=1/log(a)b
- multiplying both sides by log(a)b I got [log(a)b]^2=1
-I took the square root of both since sqrt(1) =1 and got : log (a)b=1
- applying a consequence of the definition of logarithm(log(a)a=1) I got: log(a)b=log(a)a so b=a?
Can someone kindly point me in the right direction I must have make a mistake somewhere.
I am new in the topic of logarithms.
Your help is deeply appreciated.
thank You.
Learningforcomp
 
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Re: logarithms problems,

Postby HallsofIvy » Sun Sep 13, 2020 9:02 am

Yes, in order that [tex]ax^2+ bx+ c= 0[/tex] have two equal roots, the "discriminant", [tex]b^2- 4ac[/tex], must be 0. Here your equation is [tex]x^2- log_a(b)x+ 2log_b(a)= 0[/tex] so you must have [tex](log_a(b))^2- 8log_b(a)= 0[/tex].

Since you have logarithms with two different bases, you will need to use the "change of base" formula, [tex]\log_b(a)= \frac{log_c(a)}{log_c(b)}[/tex] where c can be any positive number except 1. In particular, taking c= a, [tex]log_b(a)= \frac{1}{log_a(b)}.

So [tex](log_a(b))^2= 8log_b(a)[/tex] can be written as [tex](log_a(b))^2= \frac{8}{log_a(b)}[/tex]. Then [tex](log_a(b))^3= 8[/tex] so [tex]log_a(b)= 2[/tex] and [tex]log_b(a)= \frac{1}{2}[/tex] so that [tex]b= a^2[/tex] and [tex]a= b^{1/2}[/tex].

HallsofIvy
 
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Re: logarithms problems,

Postby Guest » Tue Sep 15, 2020 11:37 am

HallsofIvy wrote:Yes, in order that [tex]ax^2+ bx+ c= 0[/tex] have two equal roots, the "discriminant", [tex]b^2- 4ac[/tex], must be 0. Here your equation is [tex]x^2- log_a(b)x+ 2log_b(a)= 0[/tex] so you must have [tex](log_a(b))^2- 8log_b(a)= 0[/tex].

Since you have logarithms with two different bases, you will need to use the "change of base" formula, [tex]\log_b(a)= \frac{log_c(a)}{log_c(b)}[/tex] where c can be any positive number except 1. In particular, taking c= a, [tex]log_b(a)= \frac{1}{log_a(b)}.

So [tex](log_a(b))^2= 8log_b(a)[/tex] can be written as [tex](log_a(b))^2= \frac{8}{log_a(b)}[/tex]. Then [tex](log_a(b))^3= 8[/tex] so [tex]log_a(b)= 2[/tex] and [tex]log_b(a)= \frac{1}{2}[/tex] so that [tex]b= a^2[/tex] and [tex]a= b^{1/2}[/tex].


Thank you!
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