b^(log of x to the base b) = x

I don't even know solve for this myself. Could someone please show their work for this?

2. Originally Posted by bhuang

b^(log of x to the base b) = x

I don't even know solve for this myself. Could someone please show their work for this?
Given that $\displaystyle b,c \neq 0,1$ and $\displaystyle x > 0$ so the domain is satisfied

Suppose the following is true: $\displaystyle b^{log_b(x)} = x$

Take the log of both sides

$\displaystyle log_b(x)log_c(b) = log_c(x)$

From the change of base rule $\displaystyle log_b(x) = \frac{log_c(x)}{log_c(b)}$

$\displaystyle \frac{log_c(x)}{log_c(b)} \, log_c(b) = log_c(x)$

$\displaystyle log_c(b)$ cancels to give $\displaystyle log_c(x)=log_c(x)$

Therefore it is true.

3. $\displaystyle a^b=x,\ log_ax=b$

You can think of it as just "sliding" the "a" over to the underside of x,
or vice versa.

$\displaystyle y=b^{log_b(x)}$

$\displaystyle log_by=log_bx$

$\displaystyle y=x$

4. Originally Posted by bhuang

b^(log of x to the base b) = x

I don't even know solve for this myself. Could someone please show their work for this?
False.
The question should say that: $\displaystyle x>0$.

5. Originally Posted by bhuang
I'm curious- what is your definition of "$\displaystyle log_b(x)$"?
(The reason I'm curious is that what you stated is generally given as the definition! If you are using that definition, "it is true by the definition of $\displaystyle log_b(x)$" would be a perfectly good answer!)