
To calculate: .The equation for the value of x and check for extraneous solution.

Answer to Problem 34E
The required real value of x is 9 .
Explanation of Solution
Given Information:
The given equation is log63x+log6(x−1)=3 .
Formula Used:
Product rule of logarithm:
logax+logay=loga(xy)
Power rule of logarithm:
logaxp=plogax
Property of logarithms:
If loga(x)=b then x=ab
If ax2+bx+c=0 then x=−b±√b2−4ac2a .
Where a,b,c are constant and a≠0 .
Calculation:
Consider the equation log63x+log6(x−1)=3 .
Use product rule of logarithm logax+logay=loga(xy) in the above equation.
log63x(x−1)=3
Now, use the property of logarithms of the above equation.
3x(x−1)=63
Simplify the equation.
3x(x−1)=633x2−3x=2163x2−3x−216=03(x2−x−72)=0
Simplify the above equation.
3(x2−x−72)=0x2−x−72=0
Now, use quadratic formula in the above equation x2−x−72=0 .
x=−b±√b2−4ac2a
Substitute 1 for a , −1 for b and −72 for c in the above equation.
x=−(−1)±√(−1)2−4(1)(−72)2(1)
Simplify the equation.
x=−(−1)±√(−1)2−4(1)(−72)2(1)=1±√1+2882=1±√2892=1±√1722
Simplify the equation.
x=1±√1722=1±172=1+172,1−172=9,−8
Check for extraneous solution:
For x=9
log63x+log6(x−1)=3
Substitute 9 for x in the given above equation.
log63(9)+log6(9−1)=3
Simplify the equation.
log63(9)+log6(9−1)=3log627+log68=3
Use product rule of logarithm logax+logay=loga(xy) in the above equation.
log6(27)(8)=3
Simplify the equation.
log6216=3log663=3
Use power rule of logarithm logaxp=plogax of the above equation.
3log66=3
Substitute 1 for log66 in the above equation.
3(1)=33=3
For x=−8
log63x+log6(x−1)=3
Substitute −8 for x in the given above equation.
log63(−8)+log6(−8−1)=3
Simplify the equation.
log63(−8)+log6(−8−1)=3log6(−24)+log2(−9)=3
It is known that loga(−x) is not exist.
Thus, at x=−8 the given function has extraneous solution.
Hence, the required real value of x is 9 .
Chapter 6 Solutions
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