SOLUTION: Solve for x:{{{(logx)^3=logx^4}}}

Question 359742: Solve for x:
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!

This equation can be solved if we rewrite it in terms of log(x). The right side of the equation can be rewritten in terms of log(x) if we use a property of logarithms, , to move the exponent of out in front:

If you can't yet see how to solve this, perhaps a temporary variable will help. Let z = log(x). Substituting z in for log(x) we get:

To solve a 3rd degree equation like this, we want one side to be zero. So we'll subtract 4z from each side:

And then we factor. First the Greatest Common Factor (which is z):

Then we can use the difference of squares pattern, , to factor (with "a" being "z" and "b" being 2):

From the Zero Product Property we know that this (or any) product can be zero only if one of the factors is zero. So
z = 0 or z+2 = 0 or z-2 = 0
Solving these we get:
z = 0 or z = -2 or z = 2
Of course we do not care what z is. We want to know what x is. So at this point we substitute back log(x) for z:
log(x) = 0 or log(x) = -2 or log(x) = 2
If you understand what logarithms are then you may be able to figure out what x is for each of these. If not, then the next step is to rewrite these in exponential form. In general, can be rewritten as . So for your equations, with base 10 logarithms, we get:
or or
Simplifying these powers of 10 we get:
1 = x or 1/100 = x or 100 = x
So there are three solutions to your equation.

P.S. With some practice on equations like this, you will be able to work without the temporary variable. You will see how to go from

etc.
RELATED QUESTIONS
solve for x: log2x^3 - logx = log16 - logx (answered by jsmallt9)

Solve for x:... (answered by lwsshak3)

Logx-log16=log2x + logx Solve for x (answered by josmiceli,Boreal)

3^logx = 3x solve for... (answered by Edwin McCravy)

Solve for x and y if logx + logy = 4 logx + 2logy = 3 Note Bases are in base 10 (answered by josgarithmetic)

Solve for x: Logx^16 =... (answered by ankor@dixie-net.com)

solve for x: 2^logx=... (answered by bucky)

Solve for x: 3logx + 1/logx =... (answered by addingup,MathTherapy)

evaluete Logx(x+3)=4 (answered by Fombitz)