# Prove the product property of logarithms (refer to photo attached) using the properties of exponents.

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### 2 Answers

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.

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### User Comments

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

simplify:

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