Solve the following equation: log5(2x-1) + log5(2x+1)=1?

Hi,

log5(2x-1) + log5(2x+1)=1

log₅ (2x-1)(2x+1) = 1

log₅ (4x² - 1) = 1

4x² - 1 = 5¹

4x² - 1 = 5

4x² = 6

x² = 3/2

x = √(6)/2 <==ANSWER

I hope that helps!! :-)

log₅ (2x – 1) + log₅ (2x + 1) = 1 . . <- Law of logarithm: log ab = log a + log b
log₅ [(2x – 1)(2x + 1)] = 1 . . . . . . <- Difference of 2 squares: a² – b² = (a + b)(a – b)
log₅ [(2x)² – 1²] = 1
log₅ [4x² – 1] = 1 . . . . . . . . . . . . .<- log[a] x = b ⇒ x = a^b
4x² – 1 = 5¹
4x² – 1 = 5 . . . . . . . . . . . . . . . . .<- add 1 to both sides
4x² = 5 + 1
4x² = 6 . . . . . . . . . . . . . . . . . . . .<- divide both sides by 4
x² = 6/4
x² = 3/2 . . . . . . . . . . . . . . . . . . . <- square root both sides
x = ±√(3/2)

Done

First, find the slope for points (1,1) and (4,14) m(slope) = (y2 – y1) / (x2 – x1) = (14 – 1) / (4 – 1) = 13/3 Second, find slope for other four equations: Y = mx +b 5x + 6y = 19 6y = -5x + 19 y = -5/6 + 19/6 m = -5/6 5x – 6 = y y = 5x – 5 m = 5 -x + 4y = 15 4y = x + 15 y = ¼*x + 15/4 m = ¼ 4x + 5y = -1 5y = -4x -1 y = -4/5*x – 1/5 m = -4/5 Since none of the slope above match, we need to find the equation by formula y – y1 = m(x – x1) (we can pick points (1,1) or (4,14), but I rather do the easy one) Y – 1 = 13/3 (x – 1) Y – 1 = 13/3x – 13/3 Y = 13/3x – 13/3 + 1 Y = 13/3x – 10/3 Points (1, 4) and (4, 14) has the equation of y = 13/3x – 10/3

log5 (2x - 1) + log5 (2x + 1) = 1
* (x > 1/2)

log5 (2x - 1)(2x + 1) = log5 (5)
(2x - 1)(2x + 1) = 5
4x² - 1 = 5
4x² - 6 = 0
(2x - √6)(2x + √ 6) = 0
Since x > 1/2 > 0, 2x + √6 ≠ 0. Hence
2x - √6 = 0
x = √6 / 2

log5 (2x - 1) + log5 (2x + 1) = 1
log5 [(2x - 1)(2x + 1)] = 1
log5 [4x² - 1] = 1
4x² - 1 = 5
4x² = 6
x² = 1.5
x = ± 1.22474487

But, if x = -1.22474487, then the logarithm argument will be negative. Therefore, the answer is x = 1.22474487

log5(2x-1) + log5(2x+1)=1
(2x - 1)(2x + 1) = 5^1
4x^2 -2x + 2x -1 = 5
4x^2 = 6
x^2 = 6/4 = 3/2
x = +/-sqrt(3/2)

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I have to add a little something to the discussion.

logM exists only if M > 0
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This problem goes to
x = ± √(3/2)    OR    ± ½√6

Now, the original equation involves  log5(2x-1)  and  log5(2x+1).
For  x = -√(3/2)   OR   x = - ½√6:
both  (2x-1)  and  (2x+1)  are negative
so that neither  log5(2x-1)  nor  log5(2x+1) exist.

So,
x = -√(3/2)   OR   x = - ½√6  is not a solution.


Therefore,
ANSWER
x = √(3/2)   OR   x = ½√6


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Let log be log to base 5

log [ (2x - 1 ) ( 2x + 1 ) ] = 1

4 x² - 1 = 5

4 x² = 6

x ² = 3 / 2

x = ± √ (3 / 2)

If 5 is the base,then 4x^2-1=5
4x^2=6,x=1.22474
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If 5 is not the base,then 25(4x^2-1)=10
4x^2-1=2/5
4x^2=7/5,x=0.5916

Theres a rule for this for addition log a + log b= log ab so u can rewrite as log5 (2x-1)(2x+1) = log5 (4x^2-1)=1 so then you inverse log both sides with base 5 so 5^(log5(4x^2-1))=5^1 so then 4x^2-1=5 so then 4x^2-6=0 4x^2=6 so x^2=3/2 so x= +/- sqrt (3/2)