Changing the base logarithms homework

[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?!

 

[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

 

[NewScientist]

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

 

[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.

 

[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}
Your original equation is therefore:
\log_{4}x^{3}=2
Or:
x^{3}=4^{2}
which agrees with Tide's answer..

 

[TD]

which agrees with Tide's answer..

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

 

[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!

 

[arildno]

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

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

 

[TD]

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

 

[NewScientist]

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

 

[TD]

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

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.

 

[ivybond]

log_2(x) + log_4(x) = 2

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

Any ideas?!

you probably know the formula
log_a (x^n) = n log_a (x), where x&gt;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) ).

For your equation
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}

 

[ivybond]

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.

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

 

[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.