How to simplify $(3\log x) - (2\log x)$? Would this become $(\log x )^ {\frac{3}{2}}$ or would this be just $3\log x-2\log x =\log x$? If so how to get $\log x$?
I was given this question: solve for $x$ if $\log x + \log x^2 +...+ \log x^n =n(n+1)$. But, the answer to my main question will also be enough. Thank you for trying!
On
Hint: $\log(ab) = \log(a) + \log(b)$. This should be enough to simplify the sum on the left-hand side.
On
$$3\log x-2\log x$$
Here, we use two identities.
$$\log a^b = b\cdot\log a$$
$$\log_a b-\log_a c = \log_a \big(\frac{b}{c}\big)$$
Use the first identity to reach the following expression.
$$\implies \log x^3-\log x^2$$
Use the second identity to simplify.
$$\implies \log \big(\frac{x^3}{x^2}\big) \implies \boxed{\log x}$$
So no, you can’t get $(\log x)^{\frac{3}{2}}$...
For the next question, we have the following equation.
$$\log x+\log x^2+\log x^3+...+\log x^n = n\cdot(n+1)$$
Simplify the left hand side using the first identity.
$$\log x+2\cdot\log x+3\cdot\log x+...+n\cdot\log x = n\cdot(n+1)$$
$$(1+2+3+...+n)\cdot\log x = n\cdot(n+1)$$
$$\sum_{k = 1}^n k = n\cdot\frac{n+1}{2}$$
$$\biggr(\frac{n\cdot(1+n)}{2}\biggr)\cdot \log x = n\cdot(n+1)$$
$$\log x = \frac{n\cdot(n+1)}{\frac{n\cdot(1+n)}{2}}$$
$$\log x = \frac{n\cdot(n+1)\cdot 2}{n\cdot(1+n)}$$
$$\log x = 2$$
Use the definition of logarithms.
$$\log_b x = y \longleftrightarrow b^y = x$$
So, we get:
$$\boxed{x = 10^{2} = 100}$$
$3\log x - 2\log x = \log x$, just like $3y-2y=y$ no matter what $y$ is equal to.
Alternatively, you can get $$3\log x - 2\log x = \log(x^3)-\log(x^2) = \log\left(\frac{x^3}{x^2}\right) = \log x$$
but you can never under any manipulation get $$3\log x - 2\log x = \log(x)^\frac{3}{2}$$ because that equality is simply not true.
That said, for your main question, use the fact that $\log(x^n) = n\cdot \log(x)$ and your equation should become much much simpler.