Write the exponential form of the equation and solve the equation for x
Let x = 1og_2 1/8
2^x=1/8= 2^-3
x=-3
How would I express this in the
exponential form of the equation? Thanks!
Let x = 1og_2 1/8
To solve the equation, let's first rewrite the equation in exponential form:
x = log_2 (1/8)
In exponential form, logarithms can be written as:
b^x = a,
where b is the base of the logarithm and a is the value inside the logarithm.
In our equation, the base of the logarithm is 2, and the value inside the logarithm is 1/8.
We can rewrite the equation as:
2^x = 1/8
Now, let's solve the equation.
To solve for x, we need to get rid of the exponent on the left side. We can do this by taking the logarithm with base 2 of both sides of the equation:
log_2 (2^x) = log_2 (1/8)
Using the logarithmic property log_b (a^c) = c * log_b (a), we can simplify the left side:
x * log_2 (2) = log_2 (1/8)
Since log_b (b) = 1, we have:
x * 1 = log_2 (1/8)
x = log_2 (1/8)
Now, we need to evaluate the logarithm. The logarithm base 2 of a number is the exponent to which we need to raise 2 to get that number.
In this case, we need to find the exponent of 2 that gives us the value 1/8:
2^x = 1/8
2^x = 2^(-3)
By comparing the exponents, we can see that x = -3.
Therefore, the solution to the equation x = log_2 (1/8) is x = -3.