Make X the subject of 4logSqt(x) - log3x = logx^-2

T

Ted_Grendy

New member
Joined
Nov 11, 2018
Messages
36
Hello all

I have a really tough log problem. I need to make x the subject of the following:-

4logSqt(x) - log3x = logx^-2

I have had an attempt but I am not sure if this is correct and I could use some help.

1) 4logSqt(x) = 4logx^1/2 = logx^1/2 *4 = logx^2

2) logx^2 - log3x = logx^-2

3) logx^2 - log3x = log(x^2/3x) = log(x/3)

4) log(x/3) = logx^-2

5) logx^-2 = -2logx

6) log(x/3) = -2logx

7) log(x/3) + 2logx = 0 (add +2logx to both sides)

8) 2logx = logx^2

9) log(x/3) + logx^2 = 0

10) log(x/3) + logx^2 = log((x/3)*x^2) = log(x^3/3)

11) log(x^3/3) = 0

12) log(x^3/3) = logx^3 - log(1/3)

13) logx^3 - log(1/3) = 0

14) logx^3 = log(1/3)

15) logx^3 = 3logx

16) 3logx = log(1/3)

17) logx = (log(1/3))/3

This is as far as I get, can any spot any mistakes for give some advice.

Thank you.








log is to base 10
 
Ted_Grendy said:
Hello all

I have a really tough log problem. I need to make x the subject of the following:-

4logSqt(x) - log3x = logx^-2

I have had an attempt but I am not sure if this is correct and I could use some help.

1) 4logSqt(x) = 4logx^1/2 = logx^1/2 *4 = logx^2

2) logx^2 - log3x = logx^-2

3) logx^2 - log3x = log(x^2/3x) = log(x/3)

4) log(x/3) = logx^-2

5) logx^-2 = -2logx

6) log(x/3) = -2logx

7) log(x/3) + 2logx = 0 (add +2logx to both sides)

8) 2logx = logx^2

9) log(x/3) + logx^2 = 0

10) log(x/3) + logx^2 = log((x/3)*x^2) = log(x^3/3)

11) log(x^3/3) = 0

12) log(x^3/3) = logx^3 - log(1/3)

13) logx^3 - log(1/3) = 0

14) logx^3 = log(1/3)

15) logx^3 = 3logx

16) 3logx = log(1/3)

17) logx = (log(1/3))/3

This is as far as I get, can any spot any mistakes for give some advice.

Thank you.

log is to base 10
Click to expand...

Thanks for showing so much work; that's exactly what it takes to get a good answer.

You made a sign error at step 12; and with that corrected, you can get the solution in one more step after 17.

But you did a lot of unnecessary work. Look at step 4, and use the one-to-one property of the log (or, equivalently, raise 10 to the power on each side of the equation). You'll get the solution very soon after doing that.
 
I cannot see the sign error I made in step 12 Dr P

Dr.Peterson said:
Thanks for showing so much work; that's exactly what it takes to get a good answer.

You made a sign error at step 12; and with that corrected, you can get the solution in one more step after 17.

But you did a lot of unnecessary work. Look at step 4, and use the one-to-one property of the log (or, equivalently, raise 10 to the power on each side of the equation). You'll get the solution very soon after doing that.
Click to expand...


I cannot see the sign error I made in step 12.

Can you explain?
 
Ted_Grendy said:
Hello all

I have a really tough log problem. I need to make x the subject of the following:-

4logSqt(x) - log3x = logx^-2

I have had an attempt but I am not sure if this is correct and I could use some help.

1) 4logSqt(x) = 4logx^1/2 = logx^1/2 *4 = logx^2

2) logx^2 - log3x = logx^-2

3) logx^2 - log3x = log(x^2/3x) = log(x/3)

4) log(x/3) = logx^-2

5) logx^-2 = -2logx

6) log(x/3) = -2logx

7) log(x/3) + 2logx = 0 (add +2logx to both sides)

8) 2logx = logx^2

9) log(x/3) + logx^2 = 0

10) log(x/3) + logx^2 = log((x/3)*x^2) = log(x^3/3)

11) log(x^3/3) = 0

12) log(x^3/3) = logx^3 - log(1/3)

13) logx^3 - log(1/3) = 0

14) logx^3 = log(1/3)

15) logx^3 = 3logx

16) 3logx = log(1/3)

17) logx = (log(1/3))/3

This is as far as I get, can any spot any mistakes for give some advice.

Thank you.








log is to base 10
Click to expand...
After step 4 you had log(x/3) = logx^-2. Since log functions are 1 to 1, you can conclude that (x/3) = x^-2.

Now just solve (x/3) = x^-2 and make sure that you check that the (possible) solutions you get are in fact solutions to the original equations. Why do you need to do this???
 
Ted_Grendy said:
This is as far as I get, can any spot any mistakes for give some advice.
Click to expand...

Yes, also, you must put more complicated exponents and logarithm arguments inside grouping symbols to be correct:

New steps
-------------

4logSqt(x) - log(3x) = log[x^(-2)]


1) 4logSqt(x) = 4logx^(1/2) = logx^[(1/2)*4] = log(x^2) \(\displaystyle \ \ \ \ \ \ \) **

2) log(x^2) - log(3x) = log[x^(-2)]

3) log(x^2) - log(3x) = log[x^2/(3x)] = log(x/3)

4) log(x/3) = log[x^(-2)]

5) x/3 = x^(-2)

.
.
.

__________________


** The domains of 4logSqt(x) and log(x^2) are different, but you would be checking any candidates in he original equation, anyway.
 
Last edited:
Ted_Grendy said:
Hello all

I have a really tough log problem. I need to make x the subject of the following:-
...
11) log(x^3/3) = 0

12) log(x^3/3) = logx^3 - log(1/3)
Click to expand...
log(x^3/3) = logx^3 + log(1/3)

or

log(x^3/3) = logx^3 - log(3/1)

This is the sign error Dr.P was referring to in his post (#2).
 
You must log in or register to reply here.
Top