# Changing the base logarithms homework

1. Oct 3, 2005

### NewScientist

log_2(x) + log_4(x) = 2

I've tried everything i can think of, including changing the base - to no avail.

Any ideas?!

2. Oct 3, 2005

### TD

It's doable by changing base, but you can also do:

$$\log _2 x + \log _4 x = 2 \Leftrightarrow 4^{\log _2 x + \log _4 x} = 4^2 \Leftrightarrow 4^{\log _2 x} 4^{\log _4 x} = 16 \Leftrightarrow x^2 \cdot x = 16$$

3. Oct 3, 2005

### NewScientist

Can you explain those steps to me because I don't really understand

4. Oct 3, 2005

### TD

- We raise both sides as exponents of base 4.
- Then we use the fact that $a^{b + c} = a^b \cdot a^c$
- Then we can simplify the powers and logs because they are inverse operation.

Please specify what steps you have trouble with.

You can also do:

$$\log _2 x + \log _4 x = 2 \Leftrightarrow \log _4 x^2 + \log _4 x = 2 \Leftrightarrow \log _4 \left( {x \cdot x^2 } \right) = 2 \Leftrightarrow \log _4 \left( {x^3 } \right) = 2$$

And then take 4^ again of both sides.

5. Oct 3, 2005

### arildno

Here's how you can change base:
By definition, for y>0
$$y=2^{\log_{2}y}=4^{\log_{4}y}$$
Thus, from the last identity, we get:
$$\log_{2}y\log_{4}2=\log_{4}y$$
But:
$$2=4^{\frac{1}{2}}$$
thus:
$$\log_{4}2=\log_{4}4^{\frac{1}{2}}=\frac{1}{2}$$
or:
$$\log_{2}y=2\log_{4}y$$
Thus, setting y=x, we have:
$$\log_{2}x+\log_{4}x=3\log_{4}x=\log_{4}x^{3}$$
$$\log_{4}x^{3}=2$$
Or:
$$x^{3}=4^{2}$$

6. Oct 3, 2005

### TD

Hmm, I've seen a Tide here, but it's not me

7. Oct 3, 2005

### NewScientist

This is going to sound pathetic so i apologise but i do not know how these steps work!

$$\log _2 x + \log _4 x = 2 \Leftrightarrow 4^{\log _2 x + \log _4 x} = 4^2$$

$$4^{\log _2 x} 4^{\log _4 x} = 16 \Leftrightarrow x^2 \cdot x = 16$$

I'm really sorry but can you give me an anser for first principles - this isn't a requirement of the questoin but i actually want to understand this!

8. Oct 3, 2005

### arildno

Allright tiddely-doo, I made a mistake okay!!
I won't do it again..

9. Oct 3, 2005

### TD

Well, we use that:
$$\begin{gathered} a^{\log _a x} = x \hfill \\ a^{b + c} = a^b \cdot a^c \hfill \\ \end{gathered}$$

10. Oct 3, 2005

### NewScientist

i didn't know the result a^log_a(x) = x

11. Oct 3, 2005

### TD

That is because they are by definition inverse operations.

$$a^{\log _a x} = x \Leftrightarrow \log _a \left( {a^{\log _a x} } \right) = \log _a x \Leftrightarrow \log _a x \cdot \log _a a = \log _a x \Leftrightarrow \log _a x = \log _a x$$

So it is indeed correct.

12. Oct 4, 2005

### ivybond

you probably know the formula
$$log_a (x^n) = n log_a (x)$$, where $$x>0$$

There is a very useful similar formula:
$$log_{a^n} (x) = (1/n) log_a (x)$$
(can be derived from $$log_a (x) = 1/ log_x (a)$$ ).

$$log_2 (x) + log_4 (x) = 2$$
$$log_2 (x) + log_{2^2} (x) = 2$$
$$log_2 (x) + (1/2) log_2 (x) = 2$$
$$(3/2) log_2 (x) = 2$$
$$log_2 (x) = 4/3$$
$$x = 2^{4/3}$$

Last edited: Oct 4, 2005
13. Oct 4, 2005

### ivybond

TD, actually this property is just a definition of logarithm:
$${\log_a x}$$ is such a number $$n$$ that
$$a^{n} = x$$
Substituting $${\log_a x}$$ for $$n$$
$$a^{\log_a x} = x$$

Last edited: Oct 4, 2005
14. Oct 4, 2005

### TD

Which is why I said "they are by definition inverse operations".

What came after that was just a way of using logarithm properties he might know to show that it is equal.