Thread: I Am Struggling A Little Bit - Logarithmic Equations

1. I Am Struggling A Little Bit - Logarithmic Equations

I am really having trouble with logarithms.

How would I go about solving these equations?

I am struggling quite a bit with logarithms so to be honest I am not really sure where to begin. I know I must get all of the logs on one side and numbers on the other, but that is about it.

1) y=200e^(0.05t)+50 solve for t

For number 1 would I begin by moving the 50 to the other side so I would have y-50= 200e^(0.05)
Would I now take the natural log to both sides?

2) 10log(I/10-12) = 87 solve for I

For this one I believe I would begin with taking 10 to both sides to cancel out the log.

So would I have 10(I/10^-12) = 10^87? Is this correct so far? Where would I go from here?

3) y = 5log base 2 of (x-3)-4 solve for X

For this one I know I need to isolate x. So would I begin by adding 4 to both sides?
If I did that I would have y+4 = 5log base 2 of (x-3). From here I am not quite sure which direction to head.

So as you can see I am struggling quite a bit. But I am trying. I would appreciate any and all help I can get. Thank you!

2. Originally Posted by Adampff
I am really having trouble with logarithms.

How would I go about solving these equations?

I am struggling quite a bit with logarithms so to be honest I am not really sure where to begin. I know I must get all of the logs on one side and numbers on the other, but that is about it.

1) y=200e^(0.05t)+50 solve for t

For number 1 would I begin by moving the 50 to the other side so I would have y-50= 200e^(0.05)
Would I now take the natural log to both sides?

divide both sides by 200 first, then take the natural log of both sides

2) 10log(I/10-12) = 87 solve for I

For this one I believe I would begin with taking 10 to both sides to cancel out the log. no, that is not correct.

10log(I/10-12) = 87

divide both sides by 10 first ...

log(I/10-12) = 8.7

(I/10 - 12) = 10^8.7

3) y = 5log base 2 of (x-3)-4 solve for X

For this one I know I need to isolate x. So would I begin by adding 4 to both sides?
If I did that I would have y+4 = 5log base 2 of (x-3). From here I am not quite sure which direction to head.

(y+4)/5 = log_2(x-3)

2^[(y+4)/5] = x-3

...

3. Alright, thank you very much for the help!

For number 1 my final answer is...

ln(y-50/200/0.05) = t My teacher said it was ok right right it like this.

For number 2 my final answer is...

I = 10^-3.0

I got this answer like this...
10log(I/10^-12) = 87 original problem
log(I/10^-12) = 8.7 divided by 10
I/(10^-12) = 10^8.7 took 10 to both sides to get rid of log
I = 10^-3.3 multiplied by 10^-12 to both sides and added exponent

As I am sure you can see I did actually right the original problem incorrect for this one. I originally had written out my problem like this. 10log(I/10-12) = 87 solve for I, but I forgot the carrot symbol between the 10 and -12. It should actually look like this. 10log(I/10^-12) = 87 solve for I