How do I find the value of ''X'' in the problem x^-2.5=.636?????

Question

Please explain step by step…………..I’m really confused.

x^-2.5=.636

Answer 1

x^-2.5=.636
(x^-2.5) ^ (-1/2.5) =.636 ^ (-1/2.5)
x^ (-2.5 * (-1/2.5) ) =.636 ^ (-1/2.5)
x^1 =.636 ^ (-1/2.5)
x = .636 ^ (-1/2.5) ~ 1.19844

Answer 2

There's a couple of ways to do it. You'd like to get x which is also x^1, Raise both sides to the reciprocal of -5/2 or -2/5:

[x^(-5/2)]^(-2/5) = .636^(-2/5)

and note that a^(-b) = (1/a)^b for the right side

you multiply exponents:
x^1 = x = (1/.636)^(2/5)

.636 is not a very nice number so use a calculator...

You can use logs:

(-5/2)*log x = log 0.636
log x = (-2/5)log 0.636
now exponentiate:
x = exp[(-2/5)log 0.636]

If .636 were a nicer number then perhaps you could do something with powers and square root, which is 1/2 power.

Answer 3

Using a scientific calculator:

Because x is raised to a power (-2.5) take the -2.5 root of both sides. If your calc. cannot take the -2.5 root of .636, try to raise .636 to the -1/2.5 power. This is the same as taking a root.
The -2,5 root of .636 is your answer.

Answer 4

You mus use log functions. Remember that:


x^2.5=.636
log x^2.5 = log.636
2.5xlogx = log .636
log x = log .636/2.5
log x = -.07862
x = 10^(-.07862)
x = .8344

But, if the equation is:
x^(-2.5) = .636, you can find the x value in the same way
x = 1.198

Answer 5

x^-2.5=1/x^2.5=0.636

Use Logarithm on both sides and solve

LHS=Log 1 - Log x^2.5=0-2.5logx

ie -2.5Log x =log (.636)

Use log tables and solve

Answer 6

since there is an exponent, use logs ....
log(.636) = -2.5 * log(X)
log (.636) / -2.5 = log(x)

Answer 7

x^-2.5=.636
x^2.5= 1/0.636
2.5*lnx=ln1-ln(0.636)
=0-(-0.4525567156)
=0.4525567156
lnx=(0.4525567156)/2.5
=0.1810226863
x =e^(0.1810226863)
=1.198442367

therefore, x=1.198442367

i hope that this helps