SOLUTION: Solve for x: 3^2x=27^x+1 2. Solve for x: 25^x-1=125^x 3.Solve for x: 16^x-2=(1/8)^x Please help me.We didn't even learn any of this is class and I don't know to how solve it or

Algebra ->  Test -> SOLUTION: Solve for x: 3^2x=27^x+1 2. Solve for x: 25^x-1=125^x 3.Solve for x: 16^x-2=(1/8)^x Please help me.We didn't even learn any of this is class and I don't know to how solve it or      Log On


   

Question 696666: Solve for x: 3^2x=27^x+1
2. Solve for x: 25^x-1=125^x
3.Solve for x: 16^x-2=(1/8)^x
Please help me.We didn't even learn any of this is class and I don't know to how solve it or start it. Please help. Step by step procedure will be helpful. Thanks in advance!
Found 2 solutions by MathTherapy, ankor@dixie-net.com:
Answer by MathTherapy(10556) About Me  (Show Source):
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x:
3%5E%28%282x%29%29+=++27%5E%28%28x+%2B+1%29%29
Note that we can write 27 as a power of 3, 27 = 3^3, so we have
3%5E%28%282x%29%29+=++3%5E%283%28x+%2B+1%29%29
both exponents are powers of 3, therefore we can write it
2x = 3(x+1)
2x = 3x + 3
-3 - 3x - 2x
-3 = x
:
You can check this on a calc
enter: 3^(2*-3) look at the result
enter: 27(-3+1) see that the result is the same
;
:
2. Solve for x:
25%5E%28%28x-1%29%29=125%5Ex
The same situation here; 25 = 5^2 and 125 = 5^3, so we can write it
5%5E%282%28x-1%29%29=5%5E%28%283x%29%29
Both exponents are the power of the same number (5) therefore
2(x-1) = 3x
2x - 2 = 3x
-2 = 3x - 2x
-2 = x
:
Check this on your calc
enter 25^(-2-1)
enter 125^(-2)
:
:
3.Solve for x: 16^x-2=(1/8)^x
16%5E%28%28x-2%29%29=%281%2F8%29%5Ex
we can write both as powers of 2, 2^4 = 16 and recall that 1/8 = 2^-3
2%5E%284%28x-2%29%29 = 2%5E%28%28-3x%29%29
so we have
4(x-2) = -3x
4x - 8 = -3x
4x + 3x = + 8
7x = +8
x = 8/7
:
Check this
Enter: 16^((8/7)-2)
Enter: (1/8)^(8/7)