(81/256)^x= 64/27?

Answer 1

The general way to do a problem with a variable in the exponent is:

Take the natural log of both sides - because ln(a^b) = b ln a:

ln [(81/256)^x] = ln (64/27)

x [ ln (81/256)] = ln (64/27)

x = ln (64/27) / ln (81/256)

Plug that in a calculator to get x.
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Another way to do THIS specific problem:

81 = 3^4, 256 = 4^4, 64 = 4^3, 27 = 3^3

So you have:

[(3/4)^4]^x = (4/3)^3

You can rewrite the left side as (3/4)^(4x)
You can rewrite the right side as (3/4)^(-3).

Now you have:

(3/4)^(4x) = (3/4)^(-3)

since the bases are the same, you can put the exponents equal to each other

4x = -3
x = -3/4

Answer 2

81= 3^4 and 27 = 3^3

256 = 4^4 and 64 = 4^3

See a pattern?

Lets write this this way, so that this is clearer for you:

81/256 = 3^4/4^4 = (3/4)^4

And

64/27 = 4^3/3^4 = (4/3)^3

So, you have to get 2 things:

1) To invert the first fraction so that the 3 is a denominator instead of a numerator. This means that x is negative. Remember that a^-p = 1/a^p

Some examples: 2^-3 = 1/2^3
(1/4)^-1 = 4/1 = 4
(3/2)^-2 = (2/3)^2 = 4/9

2) To transform a 4 into a 3 (actually into a -3) by multiplying it by this x

So, 4x = -3, and x = -3/4.

Lets now check if this is right

[(3/4)^4]^.3/4 = (3/4)^[4*(-3/4)]

Remember that (a^b)^c = a^(bc)

So, whats the exponent? (4/1)*(-3/4) = -3

So, the result will be (3/4)^-3 = (4/3)^3 = 64/27 and all is OK

If you have more doubts, please, ask

Ana

Answer 3

taking log of both sides

x*(log(81/256)=log(64/27) but 81= 3^4 ,256=2^8 and 27= 3^3 so

x*(4log3-8log2) = 6log2-3log3 ===> x=3(2log2-log3)/4(log3-2log2)= -3/4

x= -3/4

Answer 4

x = -3/4

Answer 5

(81/256)^x=64/27
(3/4)^4x=(4/3)^3
(3/4)^4x=(3/4)^-3 b/c reciprocal
4x=-3
x=-3/4 answer

Answer 6

use the logarithms .
Remember that log n^x = x log n

so x log (81/256) = log (64/27)

you find -0.75

Answer 7

x = - (3/4)

Answer 8

81=3^4
256=2^8
=>(81/256)^x=(3^4/2^8)^x
64=2^6
27=3^3
=> (3^4/2^8)^x=(2^6)/(3^3)
=> (2^8/3^4)^(-x)=(2^6)/(3^3)
=> (2^2/3)^(-4x)=(2^2/3)^3
=> -4x=3
=> x=-3/4

Answer 9

81=3^4
256=4^4 LHS becomes
[(3/4)^4]^x since a^m/b^m=(a/b)^m
(a^m)^n =a^mn
=(3/4)^4x
RHS=(4)^3/(3)^3
=(4/3)^3
=(3/4)^-3 since a^m=1/a^-m
bases are equal so powers should be equal
4x=-3
x=-3/4