SOLUTION: Solve for x. (base is the x-3)
1+(4/3)log[x-3]4=11/3
Help please.
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Question 473602: Solve for x. (base is the x-3)
1+(4/3)log[x-3]4=11/3
Help please.
Found 2 solutions by lwsshak3, rmnavalta:
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Solve for x. (base is the x-3)
1+4/3log(x-3)4=11/3
**
1+4/3log(x-3)4=11/3
4/3log(x-3)4=11/3-1
4/3log(x-3)4=11/3-3/3=8/3
log(x-3)4=24/12=2
convert to exponential form: (base(x-3) raised to log of number(2)=number(4)
(x-3)^2=4
x^2-6x+9=4
x^2-6x+5=0
(x-5)(x-1)=0
x=5
or
x=1 (reject, (x-3)>1)
Check: For x=5
1+4/3log(x-3)4
1+4/3log(5-3)4
1+4/3log(2)4
1+(4/3)*2=8/3+1=8/3+3/3=11/3
Answer by rmnavalta(7) (Show Source): You can put this solution on YOUR website!
1. Add -1 to both sides of the equation.
2. Simplify 11/3 – 1.
3. Multiply both sides of the equation by ¾.
4. Convert logarithm to exponential form. y=log[b]x is equivalent to x=b^y.
5. Expand (x-3)^2. (x-3)^2=x^2-6x+9.
6. Add -4 to both sides.
7. The resulting equation is a quadratic equation. Solve this by factoring.
8. Use zero product property to solve for the value(s) of x. Equate each factor to 0.
x-5=0 or x-1=0
x=5 or x=1 9. The values of x are 5 or 1.
x=1 should be disregarded or is not a solution because it will yield to a negative base which is 1-3=-2. Logarithm can not have a negative base.
(ANSWER: x=5)
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