Use the identity log(ab) = log a + log b to combine the logarithms on the left side into a single term. Then take antilogarithms (just take the log away) on both sides.
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Original Statement:x - 1 + 2 + log(x) = 3Simplify:x + 1 + log(x) = 3Subtract 1:x + log(x) = 2Lambert W-Function:x = (W(100*ln(10))/(ln(10)) = 1.7555794993... (rounded up).This considered log(x) to be base 10 log (x).
True
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log1 + log2 + log3 = log(1*2*3) = log6
log644 - log64 = log6(44/4) = log611
That can't really be simplified. I can though be rewritten: Log 6 = x is another way of saying: 10x = 6
Use the identity log(ab) = log a + log b to combine the logarithms on the left side into a single term. Then take antilogarithms (just take the log away) on both sides.
No. Insert the word "minus" in place of the word "plus", and you'll have a true statement.
log(-3,2.3914850069628266…i) = 1.26ie. (-3)1÷1.26=2.3914850069628266…i
log(6) or log10(6) = 0.778 (3sf). Therefore 100.778 = 6 (if you did not understand logarithms).
Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "times", "equals".
Original Statement:x - 1 + 2 + log(x) = 3Simplify:x + 1 + log(x) = 3Subtract 1:x + log(x) = 2Lambert W-Function:x = (W(100*ln(10))/(ln(10)) = 1.7555794993... (rounded up).This considered log(x) to be base 10 log (x).
True
Unfortunately, limitations of the browser used by WA means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "equals" etc.
log(x) + log(2) = log(2)Subtract log(2) from each side:log(x) = 0x = 100 = 1