Combining logarithms with coefficients and exponents

[cscott]

Could someone please check my work?

Write as a single logarithm:

$$\frac{1}{4}[2(log_2 x + 3 log_2 y) - 3 log_2 z]$$

I got $$log_2(\frac{x^2y^6}{z^3})^\frac{1}{4}$$

$$\frac{3}{5}[\frac{1}{2}(log_2 x + 3 log_2 y) - 2(log_2 x - 4 log_2 y)]$$

I got $$log_2 (\frac{x^\frac{1}{2}y^{1.5}}{x^2} \cdot y^8)^\frac{3}{5}$$

I'd also appreciate some hints on the following:

$$a^2 + b^2 = 2ab$$
prove
$$log (\frac{a + b}{2} = \frac{1}{2}(log a + log b)$$

The part I really don't get is where the exponents go...

 

[JFo]

first part is perfect although you could get rid of the 1/4 power by looking at the third equation I gave you for the second problem

the second could be simplified further by noting that

$$\frac{a^m}{a^n} = a^{m-n}$$

$$a^m * a^n = a^{m+n}$$

$${(a^m)}^n = a^{mn}$$

for the third, note that

$$(a + b)^2 = a^2 +2ab + b^2$$

can you do the rest?

 

[iNCREDiBLE]

Work looks fine.


Suppose that
$$a^2 + b^2 = 2ab$$
<==>
$$a^2 - 2ab + b^2 = 0$$
<==>
$$(a - b)^2 = 0$$
then what can we say about a and b?

 

[JFo]

the above hint will work, but in a kind of backwards way... its easier (and more direct) to come up with an expression for (a+b)/2 and use the properties of logs

 

[iNCREDiBLE]

the above hint will work, but in a kind of backwards way... its easier (and more direct) to come up with an expression for (a+b)/2 and use the properties of logs

And by above you're referring to your own? If that is the case, I wouldn't agree with you.

 

[JFo]

no I was referring to your hint

 

[iNCREDiBLE]

no I was referring to your hint

Ok, but using my hint one don't have to use any properties of logarithms..

 

[JFo]

true, but going that route bypasses what I think the exercise is meant to do, considering the context of the first two questions.

but I'll let the OP decide what method to use.

 

[dextercioby]

$$a^{2}+b^{2}=2 ab$$

Add $2ab$ to it:

$$(a+b)^{2}=4ab$$

You can get

$$\frac{a+b}{2}=\pm (ab)^{\frac{1}{2}$$

Choose the "+" sign in the RHS,so you won't get imaginary numbers.

Daniel.

 

[JFo]

yeah...what dextercioby did was what I had in mind

 

[cscott]

first part is perfect although you could get rid of the 1/4 power by looking at the third equation I gave you for the second problem

the second could be simplified further by noting that

$$\frac{a^m}{a^n} = a^{m-n}$$

$$a^m * a^n = a^{m+n}$$

$${(a^m)}^n = a^{mn}$$

for the third, note that

$$(a + b)^2 = a^2 +2ab + b^2$$

can you do the rest?

I fixed up my first two answer, but I don't understand why you add a^2 and b^2 to the right side for the third. I think I see how I can do it now, though:

$$(a + b)^2 = 2ab$$
$$\frac{(a + b)^2}{2} = ab$$
$$2 log (\frac{a + b}{2}) = log a + log b$$
$$log (\frac{a + b}{2}) = \frac{1}{2}(log a + log b)$$

dextercioby: I see where you're going, but can you explain more how you got the last step?

 

[dextercioby]

By taking sq.root from both sides of the equation ??That's how the "+-" got there.

And your post is completely wrong.

Daniel.

 

[JFo]

I fixed up my first two answer, but I don't understand why you add a^2 and b^2 to the right side for the third. I think I see how I can do it now, though:

$$(a + b)^2 = 2ab$$
$$\frac{(a + b)^2}{2} = ab$$
$$2 log (\frac{a + b}{2}) = log a + log b$$
$$log (\frac{a + b}{2}) = \frac{1}{2}(log a + log b)$$

$$(a + b)^2 = a^2 + b^2 + 2ab$$

the above equation is true for any 2 real numbers a and b. [to verify, just multiply out (a+b)(a+b)]

They told you that $a^2 + b^2 = 2ab$, substituting this into the first equation you get

$$(a + b)^2 = 4ab$$

taking the square root of both sides, you get

$$(a+b) = 2(ab)^{\frac{1}{2}}$$

(we take the positive root, because the log of a negative number is not defined)

divide both sides by two and

$$\frac{a+b}{2} = (ab)^{\frac{1}{2}}$$

take the log of both sides and... (fill in rest)