Solve the following exponential equations:
4^(2-x)=1/2048
AAAaannndd the bot gets it wrong yet again!
4^(2-x)=1/2048
2^(4-2x) = 2^-11
4-2x = -11
x = 15/2
I apologize for the error in my previous response. You are correct that the solution to the equation 4^(2-x) = 1/2048 is x = 15/2. Thank you for bringing this to my attention.
To solve the exponential equation 4^(2-x) = 1/2048, we can start by rewriting both sides of the equation with the same base.
Since 4 is equal to 2^2, we can rewrite 4^(2-x) as (2^2)^(2-x) = 2^(4-2x). Now we have the equation 2^(4-2x) = 1/2048.
To continue solving the equation, we can write 1/2048 as a power of 2. Since 2048 is equal to 2^11 (2 x 2 x 2 ... x 2, 11 times), we can rewrite 1/2048 as 2^(-11).
Now our equation becomes 2^(4-2x) = 2^(-11).
Since the bases are the same (2), we can equate the exponents, so we have 4-2x = -11.
To solve for x, we will isolate the variable. We can start by subtracting 4 from both sides of the equation:
4 - 2x - 4 = -11 - 4
This simplifies to:
-2x = -15
Next, we divide both sides of the equation by -2 to solve for x:
(-2x) / -2 = -15 / -2
This gives us:
x = 15/2
Therefore, the solution to the exponential equation 4^(2-x) = 1/2048 is x = 15/2.
We can start by taking the logarithm of both sides with base 4, since 4 is the base of the exponential function in the equation:
log4(4^(2-x)) = log4(1/2048)
Using the power rule of logarithms, we can simplify the left side to:
(2-x)log4(4) = log4(1/2048)
Since log4(4) = 1, we can simplify further to:
2-x = log4(1/2048)
To solve for x, we can isolate it by subtracting 2 from both sides:
x = 2 - log4(1/2048)
Using the change of base formula for logarithms, we can rewrite log4(1/2048) as:
log4(1/2048) = log(1/2048) / log4
Using a calculator, we can approximate log(1/2048) to be -7.9658 and log4 to be 0.6021. Substituting these values into the equation for x, we get:
x = 2 - (-7.9658 / 0.6021) ≈ 14.9
Therefore, the solution to the equation 4^(2-x) = 1/2048 is approximately x = 14.9.