_{a}x= 1+log

_{a}(7x-10a), find x in terms of a.

If 2 log_{a}x= 1+log_{a}(7x-10a), find x in terms of a.

Ans: 2a, 5a

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

log(a)x^{2}=log(a)a+log(a)(7x-10a)

log(a)x^{2}=log(a)[a(7x-10a)]

eliminate the logs

x^{2}=a(7x-10a)

x^{2}=7ax-10a^{2}

x^{2}-7ax+10a^{2}=0

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

x-5a=0

x=5a

x-2a=0

x=2a

Hi!

AS PER ARTHUR'S HELP...

2 log_{a}x= 1+log_{a}(7x-10a)

log_{a}a=1

2log_{a}x=log_{a}x^{2}

log_{a}x^{2}=log_{a}a+log_{a}(7x-10a)

Using the rule log ab=log a + log b

log_{a}x^{2}=log_{a}[(a)(7x-10a)]

Let's eliminate the logs...

x^{2}=7ax-10a^{2}

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

x^{2}-7ax+10a^{2}=0

For the FOIL...

FIRST must be (x)(x)=x^{2}

OUTER and INNER must add up to -7ax.

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

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

Let's FOIL...

FIRST...(x)(x)=x^{2}

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

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

LAST...(-5a)(-2a)=10a^{2}

x^{2}-2ax-5ax+10a^{2}=0

x^{2}-7ax+10a^{2}=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.

Vivian, use the rule logb(xy)=logb(x)+logb(y) and you'll have it.

Hi Arthur;

I appreciate your help. However, I cannot apply this rule to (7x-10a).

You're welcome again Vivian. I like your solutions because you explain every step which is very helpful for the student.

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