Question

2^x+1+4^x=8 find the x value​

Answer 1

2x+4x=8x2x+4x=8x

Note how 4=224=22

and 8=238=23

So we can re-write this as

2x+(22)x=(23)x2x+(22)x=(23)x

Then simplify to

2x+22x=23x2x+22x=23x

To make things easier for us let’s make a substitution.

y=2xy=2x

so y+y2=y3y+y2=y3

y3−y2−y=0y3−y2−y=0

Therefore 1 possible value of y is 0. But let’s look for some more. Divide by y.

y2−y−1=0y2−y−1=0

Oooh a quadratic. Let’s solve it. Factorisation ain’t lookin to good so let’s use the good ‘ole quadratic formula. We see that

y=1+5–√2y=1+52

or y=1−5–√2y=1−52

Now we have 3 different possible values of y. Now let’s solve for x.

Solving for y=0

0=2x0=2x

The graph of y=2xy=2x has asymptote of equation y=0y=0 and so no values of x will satisfy the equation here.

Solving for other values of x

y=2xy=2x

x=log(2,y)x=log⁡(2,y)

We can then substitute the values of y into here to find the corresponding values of x. I will not write them here as they are quite complicated.

Now we substitute these values of x into 4x−2x4x−2x to get our final answer. When x=log(2,1+5–√2)x=log⁡(2,1+52) we find that this is equal to 1.

That’s the easy part done and if you’re only interested in real solutions you can stop here, for now we go into the beautiful world of complex solutions.

Here x=log(2,1−5–√2)x=log⁡(2,1−52)

Recall how eix=cosx+isinxeix=cos⁡x+isin⁡x

We need to extend this to aixaix where a is just a constant.

Hmmmm. What if there was another way to write this? Well there is.

aix=eixlnaaix=eixln⁡a

From Euler’s identity,

eixlna=cos(xlna)+isin(xlna)eixln⁡a=cos⁡(xln⁡a)+isin⁡(xln⁡a)

so aix=cos(xlna)+isin(xlna)aix=cos⁡(xln⁡a)+isin⁡(xln⁡a)

This of course reduces to Euler’s identity when a=ea=e

We need to solve 2x=1−5–√22x=1−52

2x2x can be written as 2i(−ix)2i(−ix)

Using our formula we can write this as

2i(−ix)=cos(−ixln2)+isin(−ixln2)2i(−ix)=cos⁡(−ixln⁡2)+isin⁡(−ixln⁡2)

simplifying we get

2x=cos(ixln2)−isin(ixln2)2x=cos⁡(ixln⁡2)−isin⁡(ixln⁡2)

1−5–√2=cos(ixln2)−isin(ixln2)1−52=cos⁡(ixln⁡2)−isin⁡(ixln⁡2)

This is looking rather complex now (pun unintended). Let’s use the exponential forms of cosxcos⁡x and sinxsin⁡x to simplify.

Answer 2

2^x+4^x=7 is the equated form

next step: simplification

=>2^x+2^2x=7

taking log of both sides

=>log2^x+log2^2x=log7(all to the base 2)

=>x+2x=log7 base 2

find value of log 7=2.807

=>3x=2.807

=>x=0.9356 ----------Answer

If the question is not clear this method is still correct

Please mark as brainliest and ask for any queries