How in the world will you solve for x in the equation 2^x=5?
2006-11-23 18:04:53 · 6 answers · asked by fictitiousness ;-) 2 in Science & Mathematics ➔ Mathematics
Giving you a similar analogy
10^2 = 100
so, Log 100 to base 10 = 2
Similarly, if 2^x=5
then Log 5 to base 2 = X
and that is the answer
To express it nemerically, we need to do this way
2^x = 5
Log 2^x = Log 5 (taking logariths with base 10)
or x Log 2 = Log 5
or, x = Log 5/Log2 = 0.698/0.0.30102 = 2.321928095
2006-11-23 18:09:58 · answer #1 · answered by TJ 5 · 4⤊ 0⤋
2^x=5 <=>
xln2 =ln5=>
x = ln5/ln2
2006-11-23 18:24:43 · answer #2 · answered by Nechita M 1 · 0⤊ 0⤋
you take the natural log (ln) of both sides. When there is an exponent inside of a natural log, it can be pulled outside like a constant:
ln(2^x) = ln(5)
x*ln(2) = ln(5)
x = ln(5)/ln(2)
and there you go
2006-11-23 18:08:25 · answer #3 · answered by Afternoon Delight 4 · 2⤊ 1⤋
xln2 =ln5
x = ln5/ln2
2006-11-23 18:08:41 · answer #4 · answered by Jerry M 3 · 1⤊ 1⤋
if my maths is not wrong, it should be solved like this:
2^x=5
xlg2=lg5
x=(lg5)/(lg2)
x= (punch your scientific calulator for answer)
2006-11-23 18:09:27 · answer #5 · answered by friedeggy 2 · 2⤊ 1⤋