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Logarithms

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2 Answers

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
 
 
 
 

Comments

Vivian, look at my solution. The rule is applied to log(a)a+log(a)(7x-10a) which becomes
log(a)[a(7x-10a)].
I did not think of that.  Thank you!
Hi!
AS PER ARTHUR'S HELP...
2 logax= 1+loga(7x-10a)
logaa=1
2logax=logax2
logax2=logaa+loga(7x-10a)
Using the rule log ab=log a + log b
logax2=loga[(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.
 

Comments

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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