Thread: solve each equation algebraically

1/3x^2-5/6=1

3x√ x-1 +2√ (x-1)^3=0 ......... square root symbol is over the x-1 and x-1^3

x^3-26x^3/2=27

2x^2(2x-1)^1/3=4x(2x-1)^4/3

15/2^x-1=7

3^x-1 + 3^x+1 =30

Any help on any of thse problems would be appreciated!

I can't read the first, third, fourth, fifth and sixth questions. Some brackets would be nice... LaTeX would be even nicer...



Go from here...

Hello, etran3358!

I'll guess what the last one means . . .

3^x-1+3^x+1=30

3^x/3+3.3^x=30

10/3.3^x=30

3^x=9

3^x=3^2

x=2

If my solution is hard to read and understand,
it's because I typed it the way you did ...
no parentheses, no spaces.


This is what I meant . . .

We are given: .3^{x-1} + 3^{x+1} = 30

which equals: .(3^x)/3 + 3(3^x) = 30

add: .(10/3)(3^x) = 30

Multiply by 3/10: .3^x = 9

Hence: .3^x = 3^2

Therefore: .x = 2

is #1

#3 is it's 26 x raished to the 3/2 power

again i apologize if it's hard to read/understand ... i'm new to this site

also the last problem reads as 15 ALL OVER 2^x -1 which is equal to 7

any help would be well appreciated!

i wrote these questions clearer down below...please help!!

Hello, etran3358!

You still need parentheses,
or learn to use LaTeX more accurately.

is #1.

Can you see that this looks like: .

But it might be: .

Hello, etran3358!

We have: .

Factor: . .

. . . . . . . .


Therefore:

. .

. .

. .

graded assignment ... thread closed