How to solve this log 2?
March 18th, 2010Subtract 9:
e^2x = 5
Take ln of both sides:
lne^2x = ln5
ln and e are inverse functions; therefore, they cancel each other out:
2x = ln5
Divide by 2:
x = (ln5) / 2
â–º x = 0.804718956
: )
e^2x + 9 = 14
e^2x = 5
log (base 2) e^2x = log (base 2) 5
2x [log (base 2) e] = log (base 2) 5
x = [log (base 2) 5]/2 [log (base 2) e]
***x = ½ [log (base 2) 5]/[log (base 2) e] ***.
Now we are getting someplace. To convert from base e to base 2, we use the change of base formulas log (base 2) e = ln e/ln 2 = 1/ln 2, and log (base 2) 5 = ln 5/ln 2. Now we can substitute those values into the starred equation above to get this:
x = (½)[(ln 5/ln 2)/(1/ln 2)]
x = (½)[(ln 5/ln 2)(ln 2)]
x = ½ (ln 5).
Now, we shouldn't just end it here, because we have to show that this value for x actually produces the desired result. If it does, then our calculated value for x will make the original equation true:
e^2x + 9 = 14
e^2x = 5
e^2 [½ (ln 5)] = 5
e^[(2/2)(ln 5)] = 5
e^[(1)(ln 5)] = 5
e^(ln 5) = 5
5 = 5.
Since ln 5 ≈ 1.609437912, then x = (½)(ln 5) ≈ (1.609437912)/2 ≈ 0.804718956.
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