Question

How do you solve 128^x=256^x?

Answer 1

To solve the equation 128^x = 256^x, you can take the logarithm of both sides of the equation. However, after simplification, we find that the equation does not have a solution.

Step-by-step explanation:

To solve the equation 128x = 256x, you can take the logarithm of both sides of the equation. The base of the logarithm doesn't matter as long as it is consistent on both sides. Let's use base 2, since both 128 and 256 are powers of 2.

Using the property that logb(ac) = c * logb(a), where b is the base, a is the number being raised to the power, and c is the exponent, we get:

x * log2(128) = x * log2(256)

Since the two logarithms are equal, we can cancel out x from both sides of the equation:

log2(128) = log2(256)

Now, we can solve for the logarithms:

log2(27) = log2(28)

Since the bases are the same, we can equate the exponents:

7 = 8

However, this is a contradiction, so the equation does not have a solution. There are no values of x that satisfy the equation 128x = 256x.

Answer 2

Step-by-step explanation:

128 = 2⁷

256 = 2⁸

(2⁷)^x = (2⁸)^x

remember,

(a^b)^c = a^(b×c) = (a^c)^b

(2^x)⁷ = (2^x)⁸

2^x×2^x×2^x×2^x×2^x×2^x×2^x =

= 2^x×2^x×2^x×2^x×2^x×2^x×2^x×2^x

after diving both sides by (2^x)⁷ we get

1 = 2^x

log2(1) = 0 = log2(2^x) = x

x = 0

as the logarithm of 1 for any base is always 0.