ln(x + 1) - lnx = 2
ln( (x+1)/x ) = 2
e^(ln( (x+1)/x ) ) = e²
(x+1)/x = e²
x + 1 = e²x
1 = e²x - x
1 = (e²-1)x
x = 1/(e² - 1)
e² behaves as a coefficient, so you just add/subtract it like any other number.
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f (x) = log_2 (8x)
f(x) = log_2(8) + log_2 (x)
log_2 (8) is the same as "2 to what power is 8?" 2³=8, so f(x) can be rewritten as
f(x) = 3 + log_2(x)
"ln 8x / ln 2 = ln 4x" is not true. You cannot divide the numbers after the log because they are being operated on separately. 8x/2 = 4x, but ln(8x)/ln(2) is not the same as ln(8x/2).
To begin graphing, plug in any number >0 (logs cannot be taken when the number after log is 0). Start with 1:
f(1) = 3 + log_2(1)
Any log of 1 = 0.
3 + 0 = 3, first point is at (1, 3)
Then evaluate f(2):
3 + log_2 (2)
log_(base) of (same number as base) = 1, because (base)^1 = (base)
3 + 1 = 4
(2, 4)
Connect the dots with the logarithmic curve's shape.
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lnx doesn't "start" at any value. Besides, ln0 is undefined, so I'm not sure I understand your question.
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log(x²) = 2logx
Graph logx, but every y value is doubled, so the curve appears wider.