Suppose 81^x = 64. What is 27^{x+1}?
Give me a clear explanation with the answer. Thanks! :)
clearly, x = log8164 = 1/4 log364
27^{x+1} = 27*3^3x
= 27*3^(3(1/4 log364))
= 27*3^(3/4 log364)
= 27*3^((log364) * 3/4)
= 27*64^(3/4)
= 27*2^(9/2)
= 27*16√2
= 432√2
To find the value of 27^(x+1), we can start by rewriting 81^x = 64 in terms of 27.
Since 81 is equal to 3^4 and 64 is equal to 4^3, we can rewrite the equation as:
(3^4)^x = (4^3)
Now, we can apply the power of a power property, which states that (a^m)^n = a^(m*n).
This means that we can rewrite the equation as:
3^(4x) = 4^3
Next, let's rewrite 27^(x+1) in terms of 3:
27^(x+1) = (3^3)^(x+1) = 3^(3*(x+1)) = 3^(3x + 3)
Now, since we know that 3^(4x) = 4^3, we can substitute this into the expression for 27^(x+1):
27^(x+1) = 3^(3x + 3) = (3^(4x))*(3^3) = (4^3)*(3^3) = 64 * 27
Using the property that 64 * 27 is equal to 1728, we find that 27^(x+1) = 1728.
Therefore, the value of 27^(x+1) is 1728.
To find 27^(x+1), we can use the relationship between the given equation 81^x = 64 and the base numbers 81 and 27.
First, let's express both 81 and 64 with respect to the base 27.
81 can be written as 27^2 (since 27^2 = 729) and 64 can be written as 27^(2x) (since 27^(2x) = 729^(x/2)).
Now, let's substitute these expressions into the equation and solve for x:
27^(2x) = 27^2
Now, since the bases are the same, we can equate the exponents:
2x = 2
Dividing both sides of the equation by 2, we get x = 1.
So, with x = 1, let's find 27^(x+1):
27^(x+1) = 27^(1+1) = 27^2 = 729.
Therefore, the value of 27^(x+1) is 729.