SOLUTION: We are currently working on logarithmic functions, and we are given two equations that need to be solved, but I find it somewhat difficult to reach to a logical solution. i: 4*ln*

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: We are currently working on logarithmic functions, and we are given two equations that need to be solved, but I find it somewhat difficult to reach to a logical solution. i: 4*ln*      Log On


   

Question 1026998: We are currently working on logarithmic functions, and we are given two equations that need to be solved, but I find it somewhat difficult to reach to a logical solution.
i: 4*ln*(2x-1)-ln*9=2*ln*(x-1)+2*ln*(x+1)
ii:log*(2x-5)+log*(3x+7)=4*log*2
Found 2 solutions by josgarithmetic, Theo:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
ii:
What base? Maybe better help if you say.

i:
ln%28%282x-1%29%5E4%29-ln%289%29=ln%28%28x-1%29%5E2%29%2Bln%28%28x%2B1%29%5E2%29
ln%28%282x-1%29%5E4%2F9%29=ln%28%28x-1%29%5E2%2A%28x%2B1%29%5E2%29
%282x-1%29%5E4%2F9=%28x-1%29%5E2%2A%28x%2B1%29%5E2
%282x-1%29%5E4=9%28x-1%29%5E2%28x%2B1%29%5E2
%282x-1%29%5E4=9%28x%5E2-1%29%5E2-------continue to solve this fourth-degree equation.

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
i'll do the second one first.
not sure about the first one yet.

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problem ii:

log(2x-5)+log(3x+7)=4*log(2)

log(2x-5) + log(3x+7) = log((2x-5)*(3x+7))

4*log(2) = log(2^4) = log(16)

your original equation becomes:

log((2x-5)*(3x+7)) = log(16)

this is true if and only if:

(2x-5)*(3x+7) = 16

multiply the factors out to get:

6x^2 - x - 35 = 16

subtract 16 from both sides of the equation to get:

6x^2 - x - 51 = 0

factor this quadratic equation to get:

(6x+17) * (x-3) = 0

solve for x to get:

x = -17/6 or x = 3

x = -17/6 is no good because it leads to a log of a negative number which is not allowed.

the solution is x = 3.

when x = 3, your original equation of log(2x-5)+log(3x+7)=4*log(2) becomes:

log(1) + log(16) = 4*log(2)

log(1) = 0
4 * log(2) = log(16)

equation becomes 0 + log(16) = log(16) which becomes log(16) = log(16) which confirms the solution is correct and that x = 3.

the first problem proved more difficult but i was able to solve it graphically.

it's the same procedure except i couldn't factor it algebraically which is why i resorted to solving it graphically.

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problem i:

4*ln(2x-1) - ln(9) = 2*ln(x-1) + 2*ln(x+1)

4*ln(2x-1) is equal to ln((2x-1)^4)

ln((2x-1)^4) - ln(9) is equal to ln((2x-1)^4/9)

2*ln(x-1) is equal to ln((x-1)^2)

2*ln(x+1) is equal to ln((x+1)^2)

ln((x-1)^2) + ln((x+1)^2) is equal to ln((x-1)^2*(x+1)^2)

your original equation becomes:

ln((2x-1)^4/9) = ln((x-1)^2*(x+1)^2)

this is true if and only if:

(2x-1)^4/9 = (x-1)^2*(x+1)^2

multiply both sides of that equation by 9 to get:

(2x-1)^4 = 9*(x-1)^2*(x+1)^2
subtract the expression on the right side of the equation from both sides of the equation to get:

(2x-1)^4 - 9*(x-1)^2*(x+1)^2 = 0

i graphed that equation and got the following solutions:

x = -.3203772 or x = .89180581 or x = 1.9999996.

the only valid solution was x = 1.99999996 because the other 2 solutions led to a log of a negative number which is not allowed.

your solution is that x = 1.999999568 which is very close to 2.

when i replaced x with 2 in the original equation i got the expression on the left side of the equation equal to the expression on the right side of the equation which confirmed that the solution is correct, at least to the accuracy that my calculator is capable of working with.

i would guess that the solution is x = 2.

if you round 1.999999568 to 1 or 2 decimal digits, you get x = 2.

the graphical solution to problem i is shown below.
the results shown there are rounded because that what the online software does.
i also solved it graphically using the TI-84 plus calculator.
that gave me the more detailed results i showed you above.

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