SOLUTION: How do you solve a log equations where there is x on both sides and one where there is only one x on one side? Such as the following?
{{{ log ( 3, x )^2}}} ={{{ log( 3, 4 )}}}
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-> SOLUTION: How do you solve a log equations where there is x on both sides and one where there is only one x on one side? Such as the following?
{{{ log ( 3, x )^2}}} ={{{ log( 3, 4 )}}}
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Question 30319: How do you solve a log equations where there is x on both sides and one where there is only one x on one side? Such as the following?
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Thank you very much!!! Answer by sdmmadam@yahoo.com(530) (Show Source):
You can put this solution on YOUR website! log ( 3, x )^2 =log( 3, 4 )
log(base 3)(x^2) = log(base 3)(4) implies
(x^2) = 4
(using log p = log q implies p = q of course when the base is the same )
(x^2) = 2^2
x = 2 (when the powers are the same the base are equal)
Answer:x = 2
Verification: x = 2 holds.
even oral checking is enough
log (2, (x+2))^2}}} = log (2,(3x+16))
log (x+2)^2 = log (3x+16) (for the same base 2)----(1)
which implies (x+2)^2 = (3x+16)
(using log p = log q implies p = q of course when the base is the same )
That is x^2+4x+4 = 3x+16
x^2+(4x-3x)+(4-16)= 0 (grouping like terms)
x^2+(4x-3x)-12= 0 (the product is (-12) and the difference is 1)
x^2+4x-3x-12= 0
x(x+4)-3(x+4) = 0
xp-3p =0 where p= (x+4)
p(x-3) =0
(x+4)(x-3) =0
(x+4) =0 implies x = -4
(x-3) =0 implies x = 3
Verification: x =-4 in (1) gives
LHS = log (x+2)^2 = log(-4+2)^2 = log(-2)^2 = log 4 (for the base 2)
RHS = log(3x+16) = log[3X(-4)+16]= log(-12+16) = log 4 (for the base 2) = LHS
Verification: x = 3 in (1) gives
LHS = log (x+2)^2 = log(3+2)^2 = log(5)^2 = log 25 (for the base 2)
RHS = log(3x+16) = log[3X3+16]= log(9+16) = log 25(for the base 2) = LHS
Therefore our values are correct.
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