# Solve for x: 5^2x - 16^x + 3 = 0

• nae99
Let s = 4^x and expand (1+s)^2 as a power series?I'm not sure if this works, and it's definitely beyond the scope of precalculus, but we could try to use the generalized binomial theorem to look for rational solutions at least:\begin{align*}0 &= 5^{2x} - 4 \cdot 4^x + 3 = (1+4)^{2x} - 4^{x+1} + 3 = \sum_{k=0}^\infty \binom{2x}{k} 4^k - 4^{x+1} + 3 = \sum_{k
nae99

## Homework Statement

5^2x - 4(4^x) + 8 = 5

## The Attempt at a Solution

5^2x - 16^x + 8 = 5

5^2x - 16^x + 8 - 5 = 0

4(4x) = 41 4x = 4(x+1). (This is not the same as 16x.)

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.

SammyS said:
4(4x) = 41 4x = 4(x+1). (This is not the same as 16x.)

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 don't properties of exponents

Let me check this.

You write:
5^2x - 4(4^x) + 8 = 5​
So did you mean:
$$5^2x - 4(4^x) + 8 = 5$$
or:
$$5^{2x} - 4(4^x) + 8 = 5$$
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.

I like Serena said:
Let me check this.

You write:
5^2x - 4(4^x) + 8 = 5​

So did you mean:
$$5^2x - 4(4^x) + 8 = 5$$
or:
$$5^{2x} - 4(4^x) + 8 = 5$$
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.

$$5^{2x} - 4(4^x) + 8 = 5$$

On the other hand, if it were
$$4^{2x} - 4(4^x) + 8 = 5$$
It would be easy- using the "laws of exponents". If you do not know the laws of exponents ($a^x*a^y= a^{x+ y}$, $(a^x)^y= (a^y)^x= a^{xy}$), 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:
$$5^{2x} - 4(4^x) + 8 = 5$$

An obvious solution is x = 0. Numerical solution methods home in on this solution as well.

RGV

HallsofIvy said:
On the other hand, if it were
$$4^{2x} - 4(4^x) + 8 = 5$$
It would be easy- using the "laws of exponents". If you do not know the laws of exponents ($a^x*a^y= a^{x+ y}$, $(a^x)^y= (a^y)^x= a^{xy}$), 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!

yes i am aware of those two laws but how will i apply it to:
$$5^{2x} - 4(4^x) + 8 = 5$$

nae99 said:
yes i am aware of those two laws but how will i apply it to:
$$5^{2x} - 4(4^x) + 8 = 5$$

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. )

oh ok

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. )

no i did not make any error that is what i am seeing on the paper

nae99 said:
no i did not make any error that is what i am seeing on the paper

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!

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!

ok thanks

You can solve it graphically.

$5^{2x} - 4(4^x) + 8 = 5$ is equivalent to $25^{x} - 4(4^x) + 3 = 0$

Graph $y= 25^{x} - 4(4^x) + 3$ and find the y-intercepts.

SammyS said:
You can solve it graphically.

$5^{2x} - 4(4^x) + 8 = 5$ is equivalent to $25^{x} - 4(4^x) + 3 = 0$

Graph $y= 25^{x} - 4(4^x) + 3$ and find the y-intercepts.

ok i will try that

I will suggest, once again, that you look at the problem again and make sure it is not
$$4^{2x}−4(4^x)+8=5$$
which, as I said before, would be easy.

HallsofIvy said:
I will suggest, once again, that you look at the problem again and make sure it is not
$$4^{2x}−4(4^x)+8=5$$
which, as I said before, would be easy.

yep, its not 4 it is 5

HallsofIvy said:
I will suggest, once again, that you look at the problem again and make sure it is not
$$4^{2x}−4(4^x)+8=5$$
which, as I said before, would be easy.
Hello, HoI, (Check for a PM)

While you are correct that $4^{2x}−4(4^x)+8=5$ would be a more reasonable problem to solve, the solutions to $5^{2x}−4(4^x)+8=5$ are rational, which is surprising to me.

Last edited:
SammyS said:
Hello, HoI,

While you are correct that $4^{2x}−4(4^x)+8=5$ would be a more reasonable problem to solve, the solutions to $5^{2x}−4(4^x)+8=5$ are rational, which is surprising to me.

ok then, thanks

I'm not sure if this works, and it's definitely beyond the scope of precalculus, but we could try to use the generalized binomial theorem to look for rational solutions at least:

\begin{align*} 0 &= 5^{2x} - 4 \cdot 4^x + 3 = (1+4)^{2x} - 4^{x+1} + 3 = \sum_{k=0}^\infty \binom{2x}{k} 4^k - 4^{x+1} + 3 = \sum_{k=1}^\infty \binom{2x}{k} 4^k - 4^{x+1} + 4 \\ &= 4 \left(\sum_{k=1}^\infty \binom{2x}{k} 4^{k-1} - 4^x + 1 \right). \end{align*}

(Strictly speaking the infinite series might diverge, but I think it might converge 2-adically.) But I'm not sure how to proceed now... Expand $4^x$ as a power series?

## What is the equation to solve for x in the given expression?

The equation to solve for x is 5^2x - 16^x + 3 = 0.

## What is the general approach to solve an exponential equation like this one?

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.

## Can this equation be solved algebraically?

Yes, this equation can be solved algebraically by using logarithms to solve for the variable x.

## What is the relationship between logarithms and exponential equations?

Logarithms are the inverse operations of exponential functions, meaning they can be used to solve exponential equations by "undoing" the exponent.

## Are there any special cases or restrictions when solving this type of equation?

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.

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