log2(log4x) = 1
Solve the equation
The bases are 2 and 4 respectively, I'm just not sure how to signify that on the keyboard.
Thank you so much, help is really appreciated.
and by the bases are 2 and 4 I mean they're like the small numbers where the standard would be 10
get rid of log2 by raising 2^power. That gives
log_4(x) = 2^1 = 2
now raise 4, to get
x = 4^2 = 16
thank you!
To solve the equation log2(log4x) = 1, we need to apply the properties of logarithms.
Let's start by rewriting the equation using the change of base formula. The formula states that loga(b) = logc(b) / logc(a).
Therefore, we can rewrite the equation as log4x = log2(1) / log2(4).
Since log2(1) equals 0, the equation simplifies to log4x = 0 / log2(4).
The next step is to calculate log2(4). Log2(4) represents the power to which 2 must be raised to yield 4.
As 2^2 = 4, log2(4) = 2.
Substituting this value back into the equation, we have log4x = 0 / 2.
Any number raised to the power of 0 equals 1. Therefore, we have log4x = 1.
Now, we need to convert the logarithmic equation into an exponential equation. In logarithmic form, it states that logb(x) = y if and only if b^y = x.
Hence, rewriting the equation in exponential form, we have 4^1 = x.
So the value of x is 4.
Putting it all together, the solution to the equation log2(log4x) = 1 is x = 4.