Solve for X?

Question

log(base2)(x-1) -4 = -log(base2)(x+5)

help! How do i solve this? Thank you! :)

Answer 1

Use these facts:

Fact #1: 2 raised to the power of "log(x)" = x (when the log is in base 2)

Fact #2: if two numbers in the same base are multiplied, then their logs are added. divided = subtracted.
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I'll use lg2 to represent logarithm in base 2. So that lg2(8) = 3 for example.

lg2(x-1) -4 = -lg2(x+5)

lg2(x-1) + lg2(x+5) = 4

if a = b then 2^a = 2^b

2^[lg2(x-1)+lg2(x+5)] = 2^4 = 16

We use Fact #2:
2^[lg2(x-1)] * 2^[lg2(x+5)] = 16

We use Fact #1:
(x-1) * (x+5) = 16

x^2 + 4x - 5 - 16 = 0

x^2 + 4x -21 = 0

We solve the quadratic:
x = (1/2)*[-4 +/- SQRT(16 + 84)]

x = -2 +/- 5

x = -7 is rejected because (x-1)=-8 and there is no value of lg2(-8)
[no logarithms of negative numbers -- unless we are working in complex numbers]

x = +3

x-1 = 2; lg2(2) = 1 (because 2^1 = 2)

x+5 = 8; lg2(8) = 3 (because 2^3 = 8)

1 - 4 = - 3 TRUE

QFD

Answer 2

log(base2)(x-1) -4 = -log(base2)(x+5)
log(base2)(x-1) + log(base2)(x+5) = 4
log (base2) [(x-1)(x+5)] = 4 (you know that [log a + log b = log (ab)], no matter what the base is). Again you are to know that log (baseN) y = x >>> y = N^x (N powered to x) >>>

(x-1)(x+5) = 2^4 = 16 >>> x^2 + 4x -5 = 16>>> x^2 +4x -21 = 0

x 1= 1/2 [ -4 + sqrt (16+84)] = 1/2 [-4 +sqrt (100)] = 1/2 [-4 +10] >> x1 = +3
x 1= 1/2 [ -4 - sqrt (16+84)] = 1/2 [-4 +sqrt (100)] = 1/2 [-4 -10] >> x2= -7

Answer 3

Don't quote me but I think X=4.64385618977472.

Try http://www.webmath.com/cgi-bin/gopoly.cgi?lhs=log%282%29%28x-1%29-4&ineq=leq&rhs=-log%282%29%28x%2B5%29&variable=x&back=solve.html