Prove using good format: log(nb)x=(log(b)x)/(1+log(b)n)

To prove the given equation, we need to use properties of logarithms. The equation we want to prove is:

log(nb)x = (log(b)x) / (1 + log(b)n)

Let's start by analyzing the left side of the equation and see if we can transform it into the right side. The left side is the logarithm of `x` to the base `nb`.

Remember that for any base `a` (where `a > 0` and `a ≠ 1`), and any positive numbers `M` and `N`, the following logarithmic property holds:

log(a)(MN) = log(a)M + log(a)N

Let's apply this property to `log(nb)x`:

log(nb)x = log(nb)x = log(b)x + log(b)n (using the property with M = x and N = n)

Now, we want the right side to look like this sum we have obtained. The right side of the equation is:

(log(b)x) / (1 + log(b)n)

Notice that if we only have log(b)x in the numerator, it won't be possible to separate it into two parts unless we can factor it out from a sum or difference. But we can see that 1 + log(b)n in the denominator suggests that we should rewrite it as a single logarithmic term somehow.

To achieve the format on the right side, we need to manipulate the denominator. Since the base of the logarithm is b, we think about the properties of logarithms that could give us a term that, when combined, gives us a factor with log(b)n. This is another property of logarithms:

a^(log(a)c) = c

Applying this to our base `b`, we have:

b^(log(b)n) = n

Now we can express the `1` in the denominator as b to the power of zero since any non-zero number to the power of zero is 1:

1 = b^0

Adding this to our previous step:

1 + log(b)n = b^0 + log(b)n

Now we try to combine these into a single logarithmic term using the inverse of the previous property, where log(a)(a^c) = c:

b^0 + log(b)n = log(b)(b^0) + log(b)n = log(b)[b^0 * n] (using the log property log(a)(M) + log(a)(N) = log(a)(M*N)) = log(b)[1*n] = log(b)n

Now we see that in the denominator, we have `log(b)n`, but we needed it in the format `1 + log(b)n`, suggesting that we made a slight error in our approach since we want a sum, not a product. This is where we realize that this approach won't lead directly to the desired result, because logarithms turn products into sums, not sums into products.

However, we can still make some manipulations based on the equivalence of the terms we have. Let's represent our original sum in a different way using exponents:

log(b)x + log(b)n = log(b)x * 1/log(b)n = (log(b)x) / (b^(log(b)n)) = (log(b)x) / (n)

We see that we created a fraction, but this denominator is still not what we need. The issue is that we need an addition within the logarithm, and we're currently dealing with multiplication.

This is where we realize that the initial equation provided might be incorrect or not standard, as we cannot use basic logarithm properties to derive a sum inside a logarithm from a single log term. If the equation is not a standard identity, it is important to reassess or check if there are any typos or misinterpretations in the given problem. Therefore, as it stands, the equation as given does not seem to hold true based on known logarithmic properties. A reevaluation of the problem statement is suggested for further attempts at proof.