a few problems im having trouble with

[jokerman]

1. Given that c(x) = 5(x-6)^3 + 2 find a(x) and b(x) so that (a o b)(x) = c(x)

i can solve simple equations like this but im not sure where to start on this one, if anyone could work through it showing the steps it would help me a lot.

2. Express in terms of logarithms of x, y and z: log2 5thROOT 16x^4*y^2/z^7

log2 is log with a little 2 ahead and below the log, then the fraction "16x^4*y^2/z^7" is all inside of a 5th root. let me know if you need more of a mental picture.(* = times)

3. Solve and give exact answers: 4^2x^2 * 4^6x = 1/64

the first number is 4then the power is 2x squared and its multiplied by 4^6x. since this chapter has been dealing with logarithms can i just drop the bases of 4 and get a new equation of 2x^2 * 6x = 1/64 and if so please work through it for me.

4. 5^3x = 8^x+3

the "x+3" is the power of the 8, its NOT 8^x + 3

thanks for any help with these, these are the harder problems of the chapter that im having trouble with and he hasnt showed how to do them in class so if you could show as many steps as possible that would help me understand it better. thanks again.

 

[morson]

2. If you want to be a smart alec, you could claim it already is?

\(\displaystyle \L\ log_2 (\frac{16x^4y^2}{z^7}\)^{\frac{1}{5}\ = \frac{1}{5}\ log_2 ((\frac{16x^4y^2}{z^7}\)) = \frac{1}{5}\ [ log_2 (16x^4) + log_2 (y^2) - log_2 (z^7) ]\)

Use log laws to complete it.

3. \(\displaystyle \L\ 4^{2x^2} 4^{6x} = 4^{-3}\)

So: \(\displaystyle \L\ 4^{2x^2 + 6x} = 4^{-3}\)

Giving a quadratic equation: \(\displaystyle \L\ 2x^2 + 6x + 3 = 0\)

4. \(\displaystyle \L\ 5^{3x} = 8^{x + 3}\)

Take a log to some real base \(\displaystyle a\): \(\displaystyle \L\ log_a (5^{3x}) = log_a (8^{x + 3})\)

So, \(\displaystyle \L\ 3x log_a (5) = (x + 3) log_a (8)\)

Expanding, \(\displaystyle \L\ 3x log_a(5) = x log_a (8) + 3log_a(8)\)

So: \(\displaystyle \L\ x(3log_a (5) - log_a (8)) = 3log_a (8)\)

Finish.

 

[Denis]

14. 5^3x = 8^x+3
the "x+3" is the power of the 8, its NOT 8^x + 3

Next time, use BRACKETS: 5^(3x) = 8^(x + 3)

5^3x really means 5^3 times x, or 125x

 

[jokerman]

ok i understand what you did to get the 2x^2 + 6x + 3 = 0 but im not seeing a way to easily get a solution. (2x + ?) (x + ?) = 0 . maybe im just really tired and missing something easy here but i cant see how to get a solution to x from that without a calculator.

 

[jokerman]

and on #2 you used the log laws to get in that form. what good does using them again do except put it back into the form you just got it out of? i feel dumb now

 

[morson]

but im not seeing a way to easily get a solution. (2x + ?) (x + ?) = 0

That's cool. You're probably having trouble because it doesn't factorise over the set of rational numbers. There does exist a formula to solve any equation of the form \(\displaystyle \L\ ax^2 + bx + c = 0\), \(\displaystyle a \not=\ 0\). The formula is:

\(\displaystyle \L\ x = \frac{-b + sqrt{b^2 - 4ac}}{2a}\, \frac{-b - sqrt{b^2 - 4ac}}{2a}\\)

For the quadratic you can't solve, a = 2, b = 6, c = 3, so:

\(\displaystyle \L\ x = \frac{-6 + sqrt{36 - 24}}{4}\, \frac{-6 - sqrt{36 - 24}}{4}\\)

This simplifies to \(\displaystyle \L\ x = \frac{-6 + 2sqrt{3}}{4}\, \frac{-6 - 2sqrt{3}}{4}\\)

Which simplifies to: \(\displaystyle \L\ x = \frac{-3 + sqrt{3}}{2}\, \frac{-3 - sqrt{3}}{2}\\)

Remember this formula. Scientists use it often. For example, if you throw a ball vertically in the air on earth with a constant speed, you'd probably have to solve a quadratic equation to find the time it would hit the ground again.

Your other query was regarding the my proposition that you "complete the simplification" using log laws. You got the idea that it could not be simplified any further, and that log laws would only return it to the form "I just got it out of." Recall that I initially had:

\(\displaystyle \L\ \frac{1}{5}\ [ log_2 (16x^4) + log_2 (y^2) - log_2 (z^7) ]\)

This simplifies to:

\(\displaystyle \L\ \frac{1}{5}\ [ log_2 (16) + log_2 (x^4) + 2 log_2 (y) - 7 log_2 (z) ]\)

Which finally simplifies to (after realising that \(\displaystyle log_2 (16) = log_2 (2^4)\):

\(\displaystyle \L\ \frac{1}{5}\ [ 4 + 4 log_2 (x) + 2 log_2 (y) - 7 log_2 (z) ]\)

 

[soroban]

Hello, jokerman!

1. Given: \(\displaystyle \:c(x) \:=\:5(x-6)^3\,+\,2\)
Find \(\displaystyle a(x)\) and \(\displaystyle b(x)\) so that: \(\displaystyle \,(a\,\circ\,b)(x) \:=\: c(x)\)

Some explanation . . .

Normally, we are given two functions: \(\displaystyle \,a(x)\) and \(\displaystyle b(x)\)
. . and we are asked to make a compositive function: \(\displaystyle \:c(x) \:=\a\,\circ\,b)(x)\)

This time we are given the composite function
. . and we are asked to "disassemble" it into two subfunctions.


A simpler example: .disassemble \(\displaystyle f(x) \:=\:2x + 3\:\) into \(\displaystyle g(x)\) and \(\displaystyle h(x)\)
. . so that: \(\displaystyle \:g(h(x)) \:=\:f(x)\)

Since \(\displaystyle f(x)\) has two operations: (1) multiply by 2, (2) add 3,
. . there are the two subfunctions.

First, multiply by 2: \(\displaystyle \:h(x)\:=\:2x\)
. . Then add 3:\(\displaystyle \:g(x)\:=\:x\,+\,3\)

Check: \(\displaystyle \:g(h(x)) \:=\:g(2x) \:=\:2x\, +\, 3\) . . . yes!


Back to our problem: \(\displaystyle \:c(x)\:=\:5(x\,-\,6)^3\,+\,2\)

There are four operations: \(\displaystyle \:\begin{Bmatrix}\text{subtract 6} \\ \text{cube} \\ \text{multiply by 5} \\ \text{add 2}\end{Bmatrix}\)

We can "split" the operations anywhere.. Let's use two-and-two

The inner function will be: subtract 6 and cube.
. . Hence: \(\displaystyle \:\fbox{b(x) \:=\x\,-\,6)^3}\)

The outer function will be: multiply by 5 and add 2.
. . Hence: \(\displaystyle \:\fbox{a(x) \:=\:5x\,+\,2}\)

There!

 

[morson]

Nice explanation Soroban. I have never seen such a question but thanks to you I now know how to do them.

 

[jokerman]

thanks morson and soroban, i just started learning this stuff less than a week ago so im not too good with logs yet but this has definitely helped me with them. :wink: