Logarithms

Question

If 2 log_ax= 1+log_a(7x-10a), find x in terms of a.

 

 

 

Ans: 2a, 5a

 

Answer 1

2log(a)x=1+log(a)(7x-10a)

log(a)x²=log(a)a+log(a)(7x-10a)

log(a)x²=log(a)[a(7x-10a)]

eliminate the logs

x²=a(7x-10a)

x²=7ax-10a²

x²-7ax+10a²=0

(x-5a)(x-2a)=0

x-5a=0

x=5a

x-2a=0

x=2a

 

 

 

 

Answer 2

Hi!

AS PER ARTHUR'S HELP...

2 log_ax= 1+log_a(7x-10a)

log_aa=1

2log_ax=log_ax²

log_ax²=log_aa+log_a(7x-10a)

Using the rule log ab=log a + log b

log_ax²=log_a[(a)(7x-10a)]

Let's eliminate the logs...

x²=7ax-10a²

Let's move everything to one side and set equal to zero...

x²-7ax+10a²=0

For the FOIL...

FIRST must be (x)(x)=x²

OUTER and INNER must add up to -7ax.

LAST must be (a)(a)=a² with coefficients of (5)(2) or (10)(1), and both coefficients must be negative to render the sum of -7ax and the product of +10a².

(x-5a)(x-2a)=0

Let's FOIL...

FIRST...(x)(x)=x²

OUTER...(x)(-2a)=-2ax

INNER..(-5a)(x)=-5ax

LAST...(-5a)(-2a)=10a²

x²-2ax-5ax+10a²=0

x²-7ax+10a²=0

(x-5a)(x-2a)=0

Either or both parenthetical equation(s) must equal zero...

x-5a=0

x=5a

x-2a=0

x=2a

THANK YOU AGAIN, ARTHUR.