Help Solving Logarithmic Equations

angelus2402004 · 2008-03-20 02:59:58

I need help trying to figure out how to solve these step by step please

log (2-3x) + log (3-2x) = 1.5

logbase7(1-x) - logbase7(x+2) = logbase7(x^2)

Thank You!

John E. Franklin · 2008-03-20 03:36:56

#1.) combine two added logs by multiplying the ingredients.

log [(2-3x)(3-2x)] = 1.5

Put the 10 from log base on other side of equation.

(2-3x)(3-2x) = 10^1.5

10^1.5 is 10^(3/2) or about 31.62

Expand left side.

(2)(3) - - (3x)(2x) - (3x)(3) - (2)(2x)

6 - - 6x^2 - 9x - 4x

6 + 6x^2 - 9x - 4x

6 + 6x^2 + (-9 - 4)x

6 + 6x^2 -13x

Set equal to right side again.

6x^2 - 13x + 6 = 10 ^ (3/2)

Now what ? I'd get it all equal to zero and use quadratic formula.

Subtract the right side to both sides to get zero on the right side.

6x^2 - 13x + 6 - 10^(3/2) = 0
Quadratic formula is ax^2 + bx + c = 0, must be zero.
So a = 6, b = -13 , and c = 6 - 10^(3/2)

Quadratic formula: Negative b plus or minus the square root of b squared minus 4ac all over 2a gives you 2 answers for x.
Usually, one is right and the other is sometimes thrown away, sometimes both are right.

John E. Franklin · 2008-03-20 03:47:22

- - 13 PlusOrMinusForTwoAnswers SquareRootOF[13^2 - 4(6)(6 - 10^(3/2)] all over 2(6) = x

+13 PlusOrMinus SquareRootOf{169 - (4)(6)(6) - - (4)(6)(10^[3/2])} all over 12 = x

13 PlusOrMinus SquareRootOf{169 - (4)(36) + (24)(10^[3/2])} all over 12 = x

13 PlusOr Minus SquareRootOf{169 - (2)(72) + (24)(1000^[1/2])} all over 12 = x

24 times 24 is 576, which is (3)(2)(2)(2) squared, oh forget that, put it in and back out of 1000 doesn't help.

1000 is (5)(2)(5)(2)(5)(2), so two 5's and and two 2's can come out of the 1000^(1/2).

1000^(1/2) = (5)(2) [(5)(2)]^{1/2} => 10 squareRoot(10)

13 PlusOrMinus SquareRootOf{169 - (2)(72) + (24)(10)(10^[1/2])} all over 12 = x

2 times 72 = 144, 169 - 144 is 25.

13 PlusOrMinus SquareRootOf{25 + 240(10^[1/2])} all over 12 = x

I don't know how to simplify it anymore, but if
you just use your calculator, you can get an
approximation.

x = -1.249920592 the Right Answer because I plugged it back in.

.759649851 + .740350148 = 1.5

The other answer of x = 3.416587258 is WRONG because if you plug it into original problem,
it goes negative inside the logarithms.

Last edited by John E. Franklin (2008-03-20 04:13:51)

LuisRodg · 2008-03-20 03:47:32

logbase7(1-x) - logbase7(x+2) = logbase7(x^2)

logbase7((1-x)/(x+2)) = logbase7(x^2)
(1-x)/(x+2) = x^2
x^3 + 2x^2 + x - 1 = 0
x = 0.465571231877

angelus2402004 · 2008-03-20 05:20:48

LuisRodg wrote:

logbase7(1-x) - logbase7(x+2) = logbase7(x^2)
logbase7((1-x)/(x+2)) = logbase7(x^2)
(1-x)/(x+2) = x^2
x^3 + 2x^2 + x - 1 = 0
x = 0.465571231877

How do I find x?

angelus2402004 · 2008-03-20 05:22:22

Thank You!!

John E. Franklin wrote:

- - 13 PlusOrMinusForTwoAnswers SquareRootOF[13^2 - 4(6)(6 - 10^(3/2)] all over 2(6) = x
+13 PlusOrMinus SquareRootOf{169 - (4)(6)(6) - - (4)(6)(10^[3/2])} all over 12 = x
13 PlusOrMinus SquareRootOf{169 - (4)(36) + (24)(10^[3/2])} all over 12 = x
13 PlusOr Minus SquareRootOf{169 - (2)(72) + (24)(1000^[1/2])} all over 12 = x
24 times 24 is 576, which is (3)(2)(2)(2) squared, oh forget that, put it in and back out of 1000 doesn't help.
1000 is (5)(2)(5)(2)(5)(2), so two 5's and and two 2's can come out of the 1000^(1/2).
1000^(1/2) = (5)(2) [(5)(2)]^{1/2} => 10 squareRoot(10)
13 PlusOrMinus SquareRootOf{169 - (2)(72) + (24)(10)(10^[1/2])} all over 12 = x
2 times 72 = 144, 169 - 144 is 25.
13 PlusOrMinus SquareRootOf{25 + 240(10^[1/2])} all over 12 = x
I don't know how to simplify it anymore, but if
you just use your calculator, you can get an
approximation.
x = -1.249920592 the Right Answer because I plugged it back in.
.759649851 + .740350148 = 1.5
The other answer of x = 3.416587258 is WRONG because if you plug it into original problem,
it goes negative inside the logarithms.

John E. Franklin · 2008-03-20 10:07:49

#2.) Either factor x^3 + 2x^2 + x - 1 = 0
I'm bad at factoring too.

or "depress the cubic" as learned hundreds of years ago
and somehow convert it to a 6th power equation that becomes
a quadratic equation due to missing terms...
It's a lengthy process I'm learning on the net right now.
(Nicolo Tartaglia and Girolamo Cardano)

luca-deltodesco · 2008-03-20 10:15:44

John E. Franklin wrote:

The other answer of x = 3.416587258 is WRONG because if you plug it into original problem,
it goes negative inside the logarithms.

Does it possibly work if you work through the calculations in the complex domain?

LuisRodg · 2008-03-20 11:08:13

Or you can just plug it in the TI-89 and be done with it.... which is what I did lol.

But yes, I'd someday like to find the time to teach myself how to factor all this kinds of equations. Im bad at factoring too.

John E. Franklin · 2008-03-20 15:13:47

@luca-delto I have yet to learn exponentials whose answer goes negative.
Maybe krassi would know, but he's busy I guess.

Kurre · 2008-03-21 00:23:04

hmm lets try to find ln(-1)

gives a=0 since |-1|=1

Daniel123 · 2008-03-21 03:29:09

Rather than doing all that...

Kurre · 2008-03-21 09:45:34

Daniel123 wrote:

Rather than doing all that...

uhm well yea but i wanted to be pedagogic

calculus101 · 2008-09-20 01:44:58

log4 - 2log(x+1) = logx

Math Is Fun Forum

#1 2008-03-20 02:59:58

Help Solving Logarithmic Equations

#2 2008-03-20 03:36:56

Re: Help Solving Logarithmic Equations

#3 2008-03-20 03:47:22

Re: Help Solving Logarithmic Equations

#4 2008-03-20 03:47:32

Re: Help Solving Logarithmic Equations

#5 2008-03-20 05:20:48

Re: Help Solving Logarithmic Equations

#6 2008-03-20 05:22:22

Re: Help Solving Logarithmic Equations

#7 2008-03-20 10:07:49

Re: Help Solving Logarithmic Equations

#8 2008-03-20 10:15:44

Re: Help Solving Logarithmic Equations

#9 2008-03-20 11:08:13

Re: Help Solving Logarithmic Equations

#10 2008-03-20 15:13:47

Re: Help Solving Logarithmic Equations

#11 2008-03-21 00:23:04

Re: Help Solving Logarithmic Equations

#12 2008-03-21 03:29:09

Re: Help Solving Logarithmic Equations

#13 2008-03-21 09:45:34

Re: Help Solving Logarithmic Equations

#14 2008-09-20 01:44:58

Re: Help Solving Logarithmic Equations

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