Given the product law of logarithms, prove the law of exponents.

Given the quotient law of logarithms, prove the quotient law of exponents.

tp://www.mathsisfun.com/algebra/exponent-laws.html

I will be happy to critique your work.

Well I know that a product law of logs is

log(a) + log(b) = log(ab)

and the quotient law of logs is

log(a)-log(b)=log(a/b)

But how do I change those laws, or relate it to prove the law of exponents.
I know what the law of exponents are, but I just need help proving it IN RELATION to the law of logs (as the question states)...?

To prove the law of exponents using the product law of logarithms, we start by expressing the exponential equations as logarithmic equations. Let's consider the first law of exponents, which states that for any positive real numbers a and b, and any real number x:

a^x * b^x = (ab)^x

To prove this, we can rewrite the equation in logarithmic form:

logₐ (a^x * b^x) = logₐ ((ab)^x)

Now, using the product law of logarithms, we can split the logarithm on the left side:

logₐ (a^x) + logₐ (b^x) = logₐ ((ab)^x)

Since logₐ (a^x) and logₐ (a^x) evaluate to x by the definition of logarithm, we simplify further:

x + x = logₐ ((ab)^x)

2x = logₐ ((ab)^x)

Finally, by taking both sides of the equation as exponents of base a:

(a^x * b^x)^1 = (ab)^x

Therefore, we have proved the law of exponents using the product law of logarithms.

Similarly, to prove the quotient law of exponents using the quotient law of logarithms, we start by expressing the exponential equation as a logarithmic equation. Let's consider the quotient law of exponents, which states that for any positive real numbers a and b, and any real number x:

(a^x) / (b^x) = (a/b)^x

To prove this, we can rewrite the equation in logarithmic form:

logₐ ((a^x) / (b^x)) = logₐ ((a/b)^x)

Now, using the quotient law of logarithms, we can split the logarithm on the left side:

logₐ (a^x) - logₐ (b^x) = logₐ ((a/b)^x)

Since logₐ (a^x) and logₐ (a^x) evaluate to x by the definition of logarithm, we simplify further:

x - x = logₐ ((a/b)^x)

0 = logₐ ((a/b)^x)

This means that the value inside the logarithm is 1, since any positive number raised to the power of 0 is equal to 1. Therefore:

1 = (a/b)^x

Finally, by taking both sides of the equation as exponents of base a:

(a^x) / (b^x) = (a/b)^x

Hence, we have proved the quotient law of exponents using the quotient law of logarithms.