Algebra inequalities and exponents

[Taylor_1989]

I have been stuck on two question for sometime, and would appreciate some guidance to where I am going wrong. Here are the two questions I seem to have trouble understanding.

1. 2x+5 < x-1/4
I have had numerous attempts at this equation and seem to get the answer wrong each time. The book says the ans= - 3. Here a the two methods I have tried to get the correct answer:

first attempt: 2x+5 < x-1/4 = 2x-x > -1 -5 / 4 = x > -6 / 4

second attempt: 2x+5 < x-1/4 = 4(2x+5) < 4(x-1/4) = 8x+20 < 4x - 4/4 = 8x+20= x

8x-x+20=0 = 8x-x= 20 which I then simplify to x = 20/7. As you can properly tell I am not getting the right answer could some please set me straight.

2. 4^2n-3= 16
This question I would apprectiate if someone could show how to work this out, as I seem to have no idea where to start. I would normally try and find something related to it on Google but have had no such luck. I don't really know where to look. Is there a specific name for the type of equation?

 

[azizlwl]

I have been stuck on two question for sometime, and would appreciate some guidance to where I am going wrong. Here are the two questions I seem to have trouble understanding.

1. 2x+5 < x-1/4
I have had numerous attempts at this equation and seem to get the answer wrong each time. The book says the ans= - 3. Here a the two methods I have tried to get the correct answer:

first attempt: 2x+5 < x-1/4 = 2x-x > -1 -5 / 4 = x > -6 / 4

second attempt: 2x+5 < x-1/4 = 4(2x+5) < 4(x-1/4) = 8x+20 < 4x - 4/4 = 8x+20= x

8x-x+20=0 = 8x-x= 20 which I then simplify to x = 20/7. As you can properly tell I am not getting the right answer could some please set me straight.

2. 4^2n-3= 16
This question I would apprectiate if someone could show how to work this out, as I seem to have no idea where to start. I would normally try and find something related to it on Google but have had no such luck. I don't really know where to look. Is there a specific name for the type of equation?

2x+5<x-1/4
x<-20/4-1/4
x<-21/4

4(2x+5)<4(x-1/4)
8x -4x < -20-1

 

[acabus]

Your lack of brackets or LaTeX is both annoying and confusing.

For your first question, the answer $x < -3$ only comes about if the equation is $2x + 5 < \frac{x-1}{4}$, so I'm assuming that's what it is. From your working, it's a bit confusing as to what equation you're trying to solve. Rewriting your working with brackets/LaTeX would make everything a lot clearer for us to understand.

Again, with your second question, it's ambiguous. Do you mean $4^{2n} - 3= 16$ pr $4^{2n-3} = 16$?

If it's $4^{2n-3} = 16$, then that's easy to solve.

$4^{2n-3} = 16$ and $4^2 = 16$, then $2n-3=2$.

If you meant $4^{2n} - 3= 16$, then that's a bit more complicated, have you used logarithms before?

 

[Taylor_1989]

Your lack of brackets or LaTeX is both annoying and confusing.

For your first question, the answer $x < -3$ only comes about if the equation is $2x + 5 < \frac{x-1}{4}$, so I'm assuming that's what it is. From your working, it's a bit confusing as to what equation you're trying to solve. Rewriting your working with brackets/LaTeX would make everything a lot clearer for us to understand.

Again, with your second question, it's ambiguous. Do you mean $4^{2n} - 3= 16$ pr $4^{2n-3} = 16$?

If it's $4^{2n-3} = 16$, then that's easy to solve.

$4^{2n-3} = 16$ and $4^2 = 16$, then $2n-3=2$.

If you meant $4^{2n} - 3= 16$, then that's a bit more complicated, have you used logarithms before?

I will give it another go, the reason why I haven't put latex in is because I am having trouble with my browser, seems to screw it up. I do apologize and appreciate the help.

 

[Taylor_1989]

Here is my working out with the correct latex; hopefully.


First attempt:
2x + 5 < $\frac{x-1}{4}$ $\rightarrow$ x + 5 < $\frac{-1}{4}$ $\rightarrow$

x < $\frac{-6}{4}$

I only put the first attemp in because I think my second is completely wrong. I would like to know how this equations works out to be x = -3

The second equation is: 4^2n-3= 16. I don't understand where you get the 2n-3=2. How did you come to this answer, I understand if you simplify the equation you get, 2.5 which is the answer, but I don't understand where you go the equation from. Could you explain it in a step by step.

 

[acabus]

Here is my working out with the correct latex; hopefully.First attempt:
$2x + 5 < \frac{x-1}{4} \rightarrow x + 5 < \frac{-1}{4} \rightarrow x < \frac{-6}{4}$

I only put the first attemp in because I think my second is completely wrong. I would like to know how this equations works out to be x = -3

The second equation is: $4^{2n-3} = 16$. I don't understand where you get the 2n-3=2. How did you come to this answer, I understand if you simplify the equation you get, 2.5 which is the answer, but I don't understand where you go the equation from. Could you explain it in a step by step.

For the first equation, your first step is wrong. Where you've attempted to subtract $x$ from both sides, you've actually subtracted $x$ from the left side, but only $\frac{x}{4}$ from the left side. Adding, for example, 2 to a fraction, is not the same as adding 2 to the numerator of the fraction. In fact, your second attempt is almost right, except the final bit, the "$=x$", I have no idea how you got to that. Redo it from $8x+20 < \frac{4x-4}{4}$.

For the second equation, you have that 4 to the power of something is 16, or: $4^x = 16$. Is it not obvious from this what $x$ is equal to? What power do you have to put 4 to, to get 16?

 

[Taylor_1989]

Right now I am complete lost. Could you show me how you would workout both problems, so I have something visual to look at. It would be big help. I still can't see where the 2n-3=2 comes from. I understand that 4^2=16 but the 2n-3=2 how dose it fit into the equation?

 

[CAF123]

Right now I am complete lost. Could you show me how you would workout both problems, so I have something visual to look at. It would be big help. I still can't see where the 2n-3=2 comes from. I understand that 4^2=16 but the 2n-3=2 how dose it fit into the equation?

We are not allowed to provide complete solutions. For your question, you know that $4^2 = 16.$ We have $4^{2n-3} = 16.$ So for what value of n will we get 2 as the exponent?

 

[Taylor_1989]

We are not allowed to provide complete solutions. For your question, you know that $4^2 = 16.$ We have $4^{2n-3} = 16.$ So for what value of n will we get 2 as the exponent?

Sorry I did not know. I think I get what you are saying. Am I right is working it out like this:

4^(2n-3)=16 so 4^(2n-3)=4^2, in the way I look is the 4=4 so they cancel out and you are left with: 2n-3=2 simplified is 2.5. Am I on the right train of thought.

 

[CAF123]

I wouldn't describe the 4's as 'cancelling' out as such. In general, if we have $$a^x = a^y,$$ then $x=y.$ The base here, $a$ must be the same on both sides of the equation.

 

[Taylor_1989]

I see where you are coming from now. Thank for the help. I will give another look at the first equation and see if I can post in a better way. Once aging thanks.

 

[Taylor_1989]

I have now figured out the 1st equation. I don't know how I got that mixed up, some how I did. I would like to say thanks to everyone for there input.n