How do you solve .5^x=6?

Answer 1

(1/2)^x = 6
ln (1/2)^x = ln 6
x ln (1/2) = ln 6
x (ln 1 - ln 2) = ln 6
x (0 - ln 2) = ln 6
x(-ln 2) = ln 6
x = ln 6 / (- ln 2) = -2.584962501...

Answer 2

Hey there!

Here's the answer.

.5^x=6 --> Write the problem.
(1/2)^x=6 --> Rewrite .5^x as (1/2)^x.
2^(-x)=6 --> Use the formula (1/a)^n=a^-n.
ln(2^(-x))=ln(6) --> Take the natural log on both sides of the equation.
-x*ln(2)=ln(6) --> Use the formula ln(m^n)=n*ln(m).
-x=ln(6)/ln(2) --> Divide ln(2) on both sides of the equation.
x=-(ln(6)/ln(2)) Divide -1 on both sides of the equation.

So the answer is x=-(ln(6)/ln(2)).

Hope it helps!

Answer 3

If you mean 0.5^x=6
then yes you ahve to tak the log of both sides
x(log0.5)=log6
x=log6/log0.5
x=.7782/ -.30102
x=-2.585 rounded off on each

if you mean 5^x and Not 0.5^x then the first answer is right

Answer 4

take logarithms on both sides

log 5^x = log 6
x log 5 = log 6
x = log 6/ log 5 = 1.113 = 1.11 (3 s.f.)

Answer 5

x = log(6)/log(.5) = approx = -2.585

Answer 6

xlog.5 = log 6
x=log6/log.5
x=-2.585