2 Answers | Add Yours
To prove: `log_(b)mn = log_(b)m + log_(b)n`
Given: w = `log_(b)m` , v =`log_(b)n`
Simplifying the given information, we get: m = `b^(w)` and n = `b^(v)`
Let us start with left hand side of the equation:
LHS = `log_(b)mn = log_(b) b^(w)b^(v)`
Using the properties of the exponent, we know that x^a * x^b = x^(a+b).
Using the same here,
LHS = `log_(b)b^(w+v) = (w+v) log_(b)b = w+v`
using the logarithmic property, `log_(e)e = 1`
Right hand side of the equation:
`log_(b)m+log_(b)n = w+v`
which is same as the left hand side of the equation.
Since LHS=RHS = w+v.
Hence the equality is proved.
log b m = w, log b n = v
this means that:
b^w = m and b^v = n
from this we can infer that:
b^p(random variable, does not matter) = m*n
which can also be written as:
log b mn = p
replace m and n with the equations:
b^p = b^w + b^v
b^p = b^(w+v)
since they have the same power:
p = w+v
since p = log b mn
and p = w+v
and w can be rewritten as log b m, and v as log b n,
we have p = w+v as:
log b mn = log b m + log b n
We’ve answered 318,926 questions. We can answer yours, too.Ask a question