# logarithms problems,

Logarithms problems.

### logarithms problems,

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
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.
thank You.
Learningforcomp

Posts: 7
Joined: Fri Sep 11, 2020 2:35 pm
Reputation: 0

### Re: logarithms problems,

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

Since you have logarithms with two different bases, you will need to use the "change of base" formula, $$\log_b(a)= \frac{log_c(a)}{log_c(b)}$$ where c can be any positive number except 1. In particular, taking c= a, $$log_b(a)= \frac{1}{log_a(b)}. So [tex](log_a(b))^2= 8log_b(a)$$ can be written as $$(log_a(b))^2= \frac{8}{log_a(b)}$$. Then $$(log_a(b))^3= 8$$ so $$log_a(b)= 2$$ and $$log_b(a)= \frac{1}{2}$$ so that $$b= a^2$$ and $$a= b^{1/2}$$.

HallsofIvy

Posts: 263
Joined: Sat Mar 02, 2019 9:45 am
Reputation: 98

### Re: logarithms problems,

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

Since you have logarithms with two different bases, you will need to use the "change of base" formula, $$\log_b(a)= \frac{log_c(a)}{log_c(b)}$$ where c can be any positive number except 1. In particular, taking c= a, $$log_b(a)= \frac{1}{log_a(b)}. So [tex](log_a(b))^2= 8log_b(a)$$ can be written as $$(log_a(b))^2= \frac{8}{log_a(b)}$$. Then $$(log_a(b))^3= 8$$ so $$log_a(b)= 2$$ and $$log_b(a)= \frac{1}{2}$$ so that $$b= a^2$$ and $$a= b^{1/2}$$.

Thank you!
Guest