To find: Given equation has an extraneous solution or not.
x=−2,8
Given:
An equation log4x+log4(x+6)=2 is given. …… ( 1)
Concept used:
If b,y>0 and b≠1 , then the exponential equation which is equivalent to the logarithmic equation logby=x is bx=y.
Calculation:
log4x+log4(x+6)=2[logamn=logam+logan]log4x(x+6)=2
Take exponentiation on each side with base 4.
4log4x(x+6)=42[aloga=1]x(x+6)=16x2+6x=16x2+6x−16=0
Now factorize of the above equation;
x2−8x+2x−16=0x(x−8)+2(x−8)=0(x+2)(x−8)=0
Thus,
x=−2,8
Now check the extraneous solution then put the value of x in the above-given equation (1):
Case(1): substitute the value of x=−2 in above equation (1)
log4(−2)+log4(−2+6)∞+log4(4)[logaa=1]∞+1∞[∞=infinity]
Therefore,
∞≠2
Now x=−2 is not satisfy the extraneous solution.
Case(2): Substitute the value of x=8 in above equation (1)
log4(8)+log4(8+6)log4(8)+log4(14)[logamn=logam+logan]log4112[log4112=3⋅403]3⋅403
Thus,
3⋅403≠2
Therefore, x=8 is not satisfy the extraneous solution.
Conclusion:
Hence, by the property of logarithm then an extraneous solution does not satisfy at x=−2,8.
Chapter 4 Solutions
Holt Mcdougal Larson Algebra 2: Student Edition 2012
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