- #1
nae99
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Homework Statement
5^2x - 4(4^x) + 8 = 5
Homework Equations
The Attempt at a Solution
5^2x - 16^x + 8 = 5
5^2x - 16^x + 8 - 5 = 0
SammyS said:4(4^{x}) = 4^{1} 4^{x} = 4^{(x+1)}. (This is not the same as 16^{x}.)
I suggest that you review properties of exponents.
It's difficult to work with logarithms (which are actually exponents) if you don't know how to work with exponents.
I like Serena said:Let me check this.
You write:
5^2x - 4(4^x) + 8 = 5
So did you mean:
[tex]5^2x - 4(4^x) + 8 = 5[/tex]
or:
[tex]5^{2x} - 4(4^x) + 8 = 5[/tex]
or yet something else?
Actually, I can't imagine you meant either the first or the second, because you won't be able to solve either.
You would only be able to solve the second form by trial and error, but I suspect that was not intended.
nae99 said:[tex]5^{2x} - 4(4^x) + 8 = 5[/tex]
HallsofIvy said:On the other hand, if it were
[tex]4^{2x} - 4(4^x) + 8 = 5[/tex]
It would be easy- using the "laws of exponents". If you do not know the laws of exponents ([itex]a^x*a^y= a^{x+ y}[/itex], [itex](a^x)^y= (a^y)^x= a^{xy}[/itex]), you should not be attempting a problem like this. Who ever gave you this problem clearly believes that you do know them. Learn the laws of exponents!
nae99 said:yes i am aware of those two laws but how will i apply it to:
[tex]5^{2x} - 4(4^x) + 8 = 5[/tex]
I like Serena said:You won't.
You'll only solve it numerically, but I do not think your current course is teaching you that.
(I still think you made a copying error when you typed in the problem. )
I like Serena said:You won't.
You'll only solve it numerically, but I do not think your current course is teaching you that.
(I still think you made a copying error when you typed in the problem. )
nae99 said:no i did not make any error that is what i am seeing on the paper
I like Serena said:Then the only way you'll solve it, is by trial and error (aka numerically).
Try filling in x=0, x=1, x=-1, x=2.
Make a graph.
Try and think of other values for x to try, like x=0.5.
There!
SammyS said:You can solve it graphically.
[itex]5^{2x} - 4(4^x) + 8 = 5[/itex] is equivalent to [itex]25^{x} - 4(4^x) + 3 = 0[/itex]
Graph [itex]y= 25^{x} - 4(4^x) + 3[/itex] and find the y-intercepts.
HallsofIvy said:I will suggest, once again, that you look at the problem again and make sure it is not
[tex]4^{2x}−4(4^x)+8=5[/tex]
which, as I said before, would be easy.
Hello, HoI, (Check for a PM)HallsofIvy said:I will suggest, once again, that you look at the problem again and make sure it is not
[tex]4^{2x}−4(4^x)+8=5[/tex]
which, as I said before, would be easy.
SammyS said:Hello, HoI,
While you are correct that [itex]4^{2x}−4(4^x)+8=5[/itex] would be a more reasonable problem to solve, the solutions to [itex]5^{2x}−4(4^x)+8=5[/itex] are rational, which is surprising to me.
The equation to solve for x is 5^2x - 16^x + 3 = 0.
The general approach to solving an exponential equation is to isolate the exponential term on one side of the equation and use logarithms to solve for the variable.
Yes, this equation can be solved algebraically by using logarithms to solve for the variable x.
Logarithms are the inverse operations of exponential functions, meaning they can be used to solve exponential equations by "undoing" the exponent.
Yes, when solving exponential equations, we must make sure that both sides of the equation have the same base. Additionally, if the base is a negative number, we must check for extraneous solutions.