log2(x)- log2(7)
does this equal 1?
To determine if log2(x) - log2(7) equals 1, we can use the logarithmic property known as the "difference rule." According to this rule, logb(a) - logb(b) is equivalent to logb(a/b).
Using this property, we can rewrite the expression:
log2(x) - log2(7) = log2(x/7)
Now, to check if this expression equals 1, we need to solve for x.
1 = log2(x/7)
To eliminate the logarithm, we can rewrite the equation in exponential form. In exponential form, logb(a) = c is equivalent to b^c = a.
So, we rewrite the equation:
2^1 = x/7
Simplifying further, we have:
2 = x/7
To solve for x, we multiply both sides of the equation by 7:
2 * 7 = x
x = 14
Therefore, the value of x that satisfies the equation log2(x) - log2(7) = 1 is x = 14.