Can somebody help me with this equation?

Question

I have wish to solve 5*4^(2x+1)-10*4x+1=0 but do not know how to go about it, can anybody help?

sorry this should be 5*4^(2x+1)-10*4^x+1=0

Answer 1

5*4^(2x+1) - 10*4^x + 1=0
let 4^x=y then 4^(2x)=y^2 and 4^(2x+1)=4y^2
therefore we can write
5*4Y^2-10y+1=0
i.e. 20y^2-10y+1=0
hope you can solve this quadratic equation
you wii get y=(5+sqrt5)/20 and y=(5-sqrt5)/20
substitute these values in 4^x=y.
take log on both sides
log(4^x)=logy
xlog4=logy
x=(logy)/(log4)
i.e. x={log[(5+sqrt5)/20]}/log4 and x={log[(5-sqrt5)/20]}/log4

Answer 2

I don't know how to solve:

5*4^(2x+1)-10*4x+1=0

But if you mean (as I suspect):

5 * 4^(2x + 1) - 10 * 4^x + 1 = 0

I can solve that.

5 * 4^(2x) * 4^1 - 10 * 4^x + 1 = 0

20 * 4^(2x) - 10 * 4^x + 1 = 0

Now, if you use the substitution u = 4^x, then u² = 4^(2x), and you have:

20u² - 10u + 1 = 0

Using the quadratic:

u = (10 ± √((-10)² - 4 * 1 * 20))/(2 * 20)
u = (10 ± √(100 - 80))/40
u = (10 ± √20)/40
u = (10 ± 2√5)/40
u = (5 ± √5)/20

Then re-substitute:

4^x = (5 ± √5)/20
log 4^x = log((5 ± √5)/20)
x log 4 = log((5 ± √5)/20)
x = log((5 ± √5)/20)/log 4

Ugly, but solvable.

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If your equation is really:

5*4^(2x+1)-10*4^x+1=0

Then my answer above is correct.

If your equation is really:

5*4^(2x+1)-10*4^(x+1)=0

Where 4 is raised to the x + 1 power, not 4 raised to the x power, plus 1, then the answer provided by other askers, 1/2, is correct:

5 * 4^(2x + 1) - 10 * 4^(x + 1) = 0
5 * 4^(2x + 1) = 10 * 4^(x + 1)
(2^2)^(2x + 1) = 2 * (2^2)^(x + 1)
2^(4x + 2) = 2^1 * 2^(2x + 2)
2^(4x + 2) = 2^(2x + 3)
4x + 2 = 2x + 3
2x = 1
x = 1/2

This is why parentheses are SOOOO important! Your first version of the equation was unsolvable. Your second version was solvable, but ugly. This one was easiest...but which one is the right one?

Answer 3

You can't go anywhere by trying like this...

0 = 5 * 4 ^ (2x + 1) - 10 * 4x + 1
1 = 5 * 4 ^ (2x + 1) - 10 * 4x
1/5 = 4 ^ (2x + 1) - 8x
1/5 + 8x = 4 ^ (2x + 1)
1/5 + 8x = 2 ^ (4x + 2)

We can go nowhere from here. But If we look from this aspect...

5*4^(2x+1) = 10*4x+1

10*4x+1 will always be an odd number

So we desire 5*4^(2x+1) to be a odd number

4 ^ y is odd only y = 0 with 4 ^ 0 = 1

so (2x+1) = 0

2x = -1
x = -1/2

5*4^(2x+1) = 5 * 1

5 = 10*4x+1
5 = 10* 4(-1/2) + 1
5 = 10*(-2) +1
5 = 21

This is not true neither. So this question is wrong due to the inputs you gave or you should check logaritmic equations and coplex numbers. But it has definatly got no solution in real numbers. If it makes you feel so sad, you should PM message me and I should solve it with advanced math. tecniques...

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It is more solvable with new equation...

5*4^(2x+1)-10*4^x+1=0

5*4^(2x+1) = 10*4^x+1

4^(2x+1) = 2*4^x+1

2 ^ (4x + 2) = 2 ^ (2x + 1) + 2 ^ 0

Due to logatirmic equations

4x + 2 = (2x+1) . 0

4x + 2 = 0
4x = -2
x = -1/2

Not checked , might be wrong...

Answer 4

I believe the answer is x = 1/2

If you substitute this into the equation it gives the correct answer

Solution:

5[4^(2x+1)] - 10[4^(x+1) = 0

5[4^(2x) * 4^1)] - 10[4^x * 4^1] = 0

20[4^(2x)] - 40[4^x] = 0

20[4^(2x)] = 40[4^x]

Divide both sides by 20

[4^(2x)] = 2[4^x]

Expand contents of first bracket

[4^x * 4^x] = 2[4^x]

Cancel 4^x from both sides

[4^x] = 2

The square root of 4 is 2

Therefore x = 1/2

Sustitution into the original equation proves this is the correct answer.

Answer 5

Is a trascendental equation, this means it can't be solved with pure mathematics. You would have to use computer algorithms.

No that you corrected the equation it becomes solvable:
5*4^(2x+1)-10*4^x+1=0
5*4^2x * 4 - 10*4^x +1 =0
20*4^2x - 10*4^x +1 =0
let y=4^x, this means that:
20*y^2 - 10*y +1
solve the quadratic equation for y and then solve x using logarithms, the result should be something like:

x = ln(5+-sqrt(5))/ln(4)

Answer 6

I'd probably start like this:

5*4^(2x+1)-10*4x-1=0 this is your original equation.
5(4^(2x+1))-40x-1=0 rewrite the equation so it looks less complicated and simplify
5(4^(2x+1))-5(8x)-1=0 so 5 is a common factor, which is useful, but not yet
5(4^(2x+1))-5(8x)=1 adding 1 to both sides
4^(2x+1)-8x=1/5 dividing both sides by 5 (I told you that 5 being a common factor would be useful ;-) )
4^(2x)-2x=1/20 divide both sides by 4
4^(2x)=2x+1/20 add 2x to both sides
ln(4^(2x))=ln(2x+1/20) take natural logarithm of both sides

According to http://mathworld.wolfram.com/Logarithm.html

log(a+b) simplifies to b=sqrt((a^2)-1)

so, using only the right hand side of our last equation,

1/20=sqrt((2x)^2-1) and squaring both sides, we get
1/400=(2x)^2-1 and adding 1 both sides, we get
401/400=(2x)^2 and taking square root of both sides, we get
sqrt(401/400)=2x and dividing both sides by 2, we get
(sqrt(401/400))/2=x

PHEW! That was a workout. I hope this helped!

Answer 7

you need to transpose the equation so that your left with x = blah blah. for example to place the -10 over change it to a +10 so then you have 5*4^(2x+1)*4+1 = 0+10 and so on.

Answer 8

because exponential means to multiply
we can say

5*4*(4^x)*(4^x)-10*4*4^x=0

(4^x)*(20*4^x-40)=0

as 4^x cannot be 0 then (20*4^x-40) should be

20*4^x=40

4^x=2

so

x = 1/2

I think this is the solution because answers above are illogically complicated.

Answer 9

5*4^(2x+1) = 10*4^(x+1)
4^(2x+1) = 2*4^(x+1)
4^(2x+1) = 4^(1/2)*4^(x+1)
4^(2x+1) = 4^(x+1.5)
2x+1 = x+1.5
x = 0.5

Answer 10

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