SOLUTION: please help me solve this question for the value of x (5^-x)(5^x-1)=0 this is what i have tried out so far (5^-x)(5^x-1)=0 5^[(-x)+(x-1)]=0 5^[-2x-1]=0 (-2x-1)log5=0 -2x-

Question 403413: please help me solve this question for the value of x
(5^-x)(5^x-1)=0 this is what i have tried out so far
(5^-x)(5^x-1)=0
5^[(-x)+(x-1)]=0
5^[-2x-1]=0
(-2x-1)log5=0
-2x-1=(log0/log5)
-2x=1
x= -(1/2)
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
If the equation is

then there is no solution. This equation says that the product of two powers of 5 is zero. But a product can be zero only if one of the factors is zero. And a power of 5 can never be zero! So there is no way for either factor to be zero, and, therefore, for this product to be zero. No solution.

There is flaw in your "solution"
(5^-x)(5^x-1)=0
5^[(-x)+(x-1)]=0
Good!
5^[-2x-1]=0
??? -x + x is zero, not -2x. So you should now have:
5^(-1) = 0
which simplifies to:
1/5 = 9
which is a false statement. Another indication of no solution.

If the equation is actually

then there is a solution. This equation still says a product is zero. But it is no longer a product of two powers of 5. The first factor is still a power of 5. But the second factor is now a power of 5 minus 1. The first factor still cannot be zero. But the second one can. The solution to this equation will be the solution to:

First we isolate the exponential term by adding 1 to each side:

The quick way to finish this is just to know how a power of 5 can be a 1. What exponent for 5 would result in a 1? What exponent makes every non-zero number a 1? Answer: Zero! So x=0 is the solution to this equation.

The long way to finish is to use logarithms. (If, for example, the equation was instead then we would not know what power of 5 is 2 and so we would have to use logarithms.) Find the logarithm of each side. The base of logarithm does not make a lot of difference unless you want decimal approximations of the answer. We will use base 10 (or common) logarithms:

Next we use a property of logarithms, , to move the exponent of the argument out in front. (It is this very property that is the reason we use logarithms on equations like these. Being able to move the exponent, where the variable is, out in front puts the variable in a place where we can solve for it.) Using this property we get:
x*log(5) = log(1)
Now we just divide both sides by log(5):

Now we use our calculators to find the two logarithms and divide them. You will find that log(1) = 0 (a reflection of the fact that any non-zero number to the zero power is 1) which makes the whole fraction zero. So x = 0 is the solution this way, too!
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