Can anyone solve this logarithmic equation? log2(x+1) + log2(x -1) = 3?

Question

Can anyone solve this logarithmic equation?
log2(x+1) + log2(x -1) = 3

Answer 1

x= 3 or -3

when you add two logs together you can combine them by multiplying so you get
log2(x+1)(x-1)=3

then you do 2^ to both sides to get rid of logs
(x+1)(x-1)=2^3

this becomes x^2-1=8
then x^2 = 9
x= 3 or -3

Answer 2

log[base 2](x + 1) + log[base 2](x - 1) = 3

First, use the log identity to combine the sum of logs into the log of a product.

log[base 2] [ (x + 1)(x - 1) ] = 3

Convert to exponential form,

2^3 = (x + 1)(x - 1)

Expand and simplify,

8 = x^2 - 1
9 = x^2

Therefore, our potential solutions are x = {-3, 3}.

With logarithmic equations, you have to check for extraneous solutions by testing each value into the original equation.

Let x = -3: Then
LHS = log[base 2](x + 1) + log[base 2](x - 1)
= log[base 2](-2) + ....
OOPS! We cannot take the log of a negative number.
Reject x = -3.

If x = 3, we should be okay.

Therefore, our only solution is x = 3.

Answer 3

Take logs as base 2
log (x + 1) + log(x - 1) = 3
log [ (x + 1).(x - 1)] = 3
(x + 1).(x - 1) = 2³ = 8
x² - 1 = 8
x² = 9
x = 3 (taking +ve value for x)

Answer 4

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RE:
Can anyone solve this logarithmic equation? log2(x+1) + log2(x -1) = 3?
Can anyone solve this logarithmic equation?
log2(x+1) + log2(x -1) = 3

Answer 5

Log2 X 1

Answer 6

log a (m) + log a (n) = log a (mn)
So
log2(x^2 - 1) = 3
2^3 = x^2 - 1
8 = x^2 - 1
x^2 = 9
x = +3 or -3

Hope this helps.

Answer 7

log2(x+1) + log2(x -1) = 3
log2(x+1)(x -1) = 3
log2(x^2-1) = 3
y = x^n
logya = n
2^3 = x^2-1
9 = x^2
+/-3 = x
-3 is rejected because there are no (-)ve values
Hence, 3 is the answer

Answer 8

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if you mean: log(6x^2) - log(2x) = 1 then log(6x^2 / (2x)) = 1 ← quotient rule log(3x) = 1 3x = 10 x = 10/3 ♣♦