Hi!
AS PER ARTHUR'S HELP...
2 log_{a}x= 1+log_{a}(7x10a)
log_{a}a=1
2log_{a}x=log_{a}x^{2}
log_{a}x^{2}=log_{a}a+log_{a}(7x10a)
Using the rule log ab=log a + log b
log_{a}x^{2}=log_{a}[(a)(7x10a)]
Let's eliminate the logs...
x^{2}=7ax10a^{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}.
(x5a)(x2a)=0
Let's FOIL...
FIRST...(x)(x)=x^{2}
OUTER...(x)(2a)=2ax
INNER..(5a)(x)=5ax
LAST...(5a)(2a)=10a^{2}
x^{2}2ax5ax+10a^{2}=0
x^{2}7ax+10a^{2}=0
(x5a)(x2a)=0
Either or both parenthetical equation(s) must equal zero...
x5a=0
x=5a
x2a=0
x=2a
THANK YOU AGAIN, ARTHUR.
2/4/2014

Vivian L.
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