Expert: Sherman D. - 4/22/2005

Question
solve the equations. Please show me how you found the answer step by step.
1. (1/27)^2x+1 = 9^3x-6

2. 5^2x-1 = 15^1-x

3. log6(6is the base) x + log6(base)(x-5)=1

4. log(x+3)+logx=1

5. e^x+1 = 10

6. log64(base) x=1/3

7. lnx + 3ln2 = ln2/x

8. 10^2x-3 = 17";  

Answer
1.)
(1/27)^(2x + 1) = 9^(3x - 6)
(3^(-3))^(2x + 1) = (3^2)^(3x - 6)
3^(-3(2x + 1)) = 3^(2(3x - 6))
3^(-6x - 3) = 3^(6x - 12)
-6x - 3 = 6x - 12
-12x = -9
x = (3/4)

there are some complex values, which you will find at www.quickmath.com, but this is what i got.

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2.)
5^(2x - 1) = 15^(1 - x)
unless you typed this wrong. At quickmath, it gives me
x = (log(3) + 2log(5))/(log(3) + 3log(5))

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3.)
log6(6is the base) x + log6(base)(x-5)=1
log(6)x + log(6)(x - 5) = 1
log(6)(x(x - 5)) = 1
log(6)(x^2 - 5x) = 1
6^1 = x^2 - 5x
6 = x^2 - 5x
x^2 - 5x - 6 = 0
(x - 6)(x + 1) = 0
x = 6 or -1

Since you can't have log(-n), the answer is x = 6.


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4.)
log(x + 3) + log(x) = 1
log(x(x + 3)) = 1
log(x^2 + 3x) = 1
inverse log both sides
x^2 + 3x = e
x^2 + 3x - e = 0
x = (-b ± sqrt(b^2 - 4ac))/2a
x = (-3 ± sqrt(3^2 - 4(1)(-e)))/2(1)
x = (-3 ± sqrt(9 + 4e))/2
x = (1/2)(-3 ± sqrt(9 + 4e))
x = .728964


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5.)
e^(x + 1) = 10
log(x + 1)10 = e
x + 1 = (log 10)/(log(e))
x + 1 = 1/(log(e))
x = (1/(log(e))) - 1
x = about 1.303

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6.)
log64(base) x=1/3
If by this you meant
log(base 64)x = (1/3)
(64)^(1/3) = x
x = 4

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7.)
lnx + 3ln2 = ln(2/x)
ln(x) + ln(2^3) = ln(2/x)
ln(x) + ln(8) = ln(2/x)
ln(8x) = ln(2/x)
8x = (2/x)
8x^2 = 2
8x^2 - 2 = 0
2(4x^2 - 1) = 0
2(2x + 1)(2x - 1) = 0
x = ±(1/2)

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8.)
10^2x-3 = 17";
What did you mean by ";