(3^x)-4*(3^-x)=0 Asking for some assistance.

R

RolandHP

New member
Joined
Oct 18, 2017
Messages
4
Apologies in advance if I posted this at the wrong place, this is my first time here. It seemed to me this falls under the category beginner algebra since I saw nothing else fitting for this.


I'm asking for some help with these two tasks:
a) (3^x)-4*(3^-x)=0
b) (4^x)+1000*(4^-x)-110=0


I suppose I can manage on my own once the task a is cleared up, so there is no need to solve b i think.


Thanks in advance.
 
RolandHP said:
Apologies in advance if I posted this at the wrong place, this is my first time here. It seemed to me this falls under the category beginner algebra since I saw nothing else fitting for this.


I'm asking for some help with these two tasks:
a) (3^x)-4*(3^-x)=0
b) (4^x)+1000*(4^-x)-110=0


I suppose I can manage on my own once the task a is cleared up, so there is no need to solve b i think.


Thanks in advance.
Click to expand...
Assuming you need to solve for 'x':
a) (3^x)-4*(3^-x)=0

substitute u = 3^x → u - 4/u = 0 → u^2 - 4 = 0 → solve for u → solve for 'x'
 
Another approach is to use properties of exponents and logarithms.

3^x - 4*3^(-x) = 0

3^x = 4/3^x

Multiply each side by 3^x

3^(2x) = 4

Now use logarithms and simplify, to find that x is a ratio of two natural logs. :cool:
 
RolandHP said:
a) (3^x)-4*(3^-x)=0
b) (4^x)+1000*(4^-x)-110=0
Click to expand...
No instructions were provided. I'll assume you're supposed to be "solving" the equations.

There is a standard trick for equations of this sort. I'll show the trick for (b), because (as shown elsewhere) there are other methods for (a).

You have this:

. . . .\(\displaystyle 4^x\, +\, 1000\, \cdot\, 4^{-x}\, -\, 110\, =\, 0\)

When you have terms like this, where they're the same except for a "minus" on one of the powers, then the trick is to multiply through by another copy of the positive-power term. In this case, that would be 4^x:

. . . .\(\displaystyle 4^x\, \left(4^x\,+\, 1000\, \cdot\, 4^{-x}\, -\, 110\right)\, =\, 4^x\, (0)\)

The bases being the same, you can add (and subtract) powers to get:

. . . .\(\displaystyle 4^{2x}\, +\, 1000\, \cdot\, 4^0\, -\, 110\,\cdot\, 4^x\, =\, 0\)

. . . .\(\displaystyle \left(4^x\right)^2\, -\, 110\, \cdot\, 4^2\, +\, 1000\, =\, 0\)

This is a quadratic "in" 4^x. That is, you can treat this like a quadratic, and solve this like a quadratic. Are there factors of +1000 that add to -110? Yes; -100 and -10. So factor:

. . . .\(\displaystyle (4^x\, -\, 100)\, (4^x\, -\, 10)\, =\, 0\)

Set each factor equal to zero, solve the resulting exponential equations, etc, etc. ;)
 
You must log in or register to reply here.
Top