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5log8X=log8 16
{5x+y=11
{2x-z=0
{y+z=5

2007-05-31 02:43:43 · 4 answers · asked by Pinky fan site for 360 fans 1 in Science & Mathematics Mathematics

4 answers

If your first equation is log base 8, then move the 5 from in front of the log to the exponent of the x term.
log8 (x^5) = log8 (16)
x^5 = 16
x = 16 ^ (1/5) (or fifth root of 16)
x = 1.7411 (decimal approximation)

Now, I'm assuming that the next three equations are a system of three equations that you need to solve for x. The numbers in the parentheses indicated an equation number for references purposes only.

5x + y = 11 (1)
2x - z = 0 (2)
y + z = 5 (3)

First add equations (2) and (3)
2x - z = 0 (2)
y + z = 5 (3)
--------------------
2x + y = 5 (4)

Next multiply equation (4) by -1 and add to equation (1).
5x + y = 11 (1)
-2x - y = -5 (4)
----------------------
3x = 6 (5) (next divide both sides by 3)
x = 2

Good luck.

2007-05-31 06:22:07 · answer #1 · answered by CV 2 · 0 0

1) x = 16/5

2)x = 11-y/5

3) x= z/2

4) No x term.

2007-05-31 02:47:53 · answer #2 · answered by Lady Geologist 7 · 0 0

1-x = 16/5

2-x = 11 y/5

I dont know 3 and 4 sry

2007-05-31 02:52:09 · answer #3 · answered by La Pieta 2 · 0 0

log3(x - a million) = 4 and log3(x - a million) = log34 log3(x - a million) = 4 ................(a million) log3(x - a million) = log34 ..........(2) subtitut a million to 2 we get log3(x - a million) = log34 4 = log34 log 10^4 = log34 10000 = 34 ??????? iam sori foreheadi think of its a incorrect issue.

2016-12-30 08:44:39 · answer #4 · answered by Anonymous · 0 0

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