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# 4^2^x + 2^2^x =56

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4^2^x + 2^2^x =56

how do we solve this equation ?

Step by step please

Jun 21, 2017

#1
+3

Let y = 2x then we have 4y + 2y = 56.      (And x = log2y)

Now 4y → (22)y → (2y)2

Let z = 2y, so now we have z2 + z = 56 or z2 + z - 56 = 0.   (And y = log2z)

This factorises as. (z + 8)(z - 7) = 0

Hence: z = 7,  y = log27, x = log2(log27)

We can't have logs of negative numbers (as long as we aren't considering complex numbers) so this is the only solution.

Jun 21, 2017

#1
+3

Let y = 2x then we have 4y + 2y = 56.      (And x = log2y)

Now 4y → (22)y → (2y)2

Let z = 2y, so now we have z2 + z = 56 or z2 + z - 56 = 0.   (And y = log2z)

This factorises as. (z + 8)(z - 7) = 0

Hence: z = 7,  y = log27, x = log2(log27)

We can't have logs of negative numbers (as long as we aren't considering complex numbers) so this is the only solution.

Alan Jun 21, 2017
#2
+1

i did ur result in the calculator but it did not give the RHS (56) what the poblem ? Guest Jun 21, 2017
#3
+2

4^(2x)  + 2^(2x) =  56

Write  as

(2^2)^(2x) +  2^(2x)  =  56

(2^(2x))^2  +  2^(2x)  = 56   subtract 56 from both sides

(2^(2x))^2  + 2^(2x)  - 56  = 0

Let z  = 2^(2x)   and we have that

z^2  + z  -  56  = 0        factor

(z + 8)  ( z - 7)  = 0

Setting each factor to 0  and solving for z we have that  z = -8  or z  =7

Since an exponential cannot be negative,  z  =  7

So.....2^(2x)  = 7          take the log of both sides

log 2^(2x)  = log 7       and we can write

(2x) * log 2  = log 7      divide both sides by log2

2x =  log 7 / [  log 2 ]

Putting this into the original equation we have that

4^(log 7 / log 2 )  +  2^(log 7 / log 2)    evaluates to 56   Jun 21, 2017
#4
+1

Thank u u gave the right answer thanks very much now I know how this can be solved wish u the best luck in math knowledge

Guest Jun 21, 2017
#5
+1

Could u tell me please why Alan solution did not work ? I do not see a problem in it maybe u know where is the mistake :)

Guest Jun 21, 2017
#6
+2

I could be wrong but it looks like Alan interpreted the question this way:     4^(2^x) + 2^(2^x)  =  56

And CPhill did this:     (4^2)^x + (2^2)^x  =  56 hectictar  Jun 22, 2017
#7
+1

In Guest #1 #2 calculator he/she put the terms as (42)log2log27 + (22)log2log27  but should have put 4log27 + 2log27 to maintain the same interpretation (noting that 2log2log27 is the same as log27).

Also see GingerAle's post below.

Alan  Jun 22, 2017
edited by Alan  Jun 22, 2017
edited by Alan  Jun 22, 2017
edited by Alan  Jun 22, 2017
#8
+2

Actually, they are not interpreted same. Each interpretation gives a different solution for x.

As Hectictar pointed out:

Sir Alan’s interpretation is: 4^(2^x) + 2^(2^x)  =  56

And Sir  CPhill interpretation is:     (4^2)^x + (2^2)^x  =  56

Sir Alan’s is interpretation is exactly the same as the asked question. (This follows power-tower convention, where stacked powers are multiplied from right to left or from the top down, and the resultant product becomes the exponent to the base number.)

Sir CPhill’s solution starts with the base then increments the value by each ascending power.

The solutions for these interpretations give different results for the value of x.

-----

$$\text{For Sir Alan's solution}\\ x = log_2 (log_2(7))\\ x \approx 1.4892114692\\ \text{ }\\ \text{For Sir CPhill’s solution}\\ 2x = \dfrac {log (7) } { log (2)}\\ x= \dfrac {log (7)/ log (2)}{(2)}\\ x\approx 1.4036774610\\$$
-----

Sir Alan notes the calculator input error of the questioner; however, the questioner may have correctly entered this into the site calculator

(4^2^log(log(7, 2), 2)) + (2^2^log(log(7 , 2), 2) )

The calculator would interpret it as

(4^2)^(log(log(7, 2), 2)) + (2^2)^log(log(7 , 2), 2))

The site’s calculator does NOT follow hierarchal convention in interpreting stacked powers.

Here is an elaboration on this issue:

Use of parenthesis to, in this case, force proper hierarchal convention

(4^(2^log(log(7, 2), 2)) + (2^(2^log(log(7 , 2), 2)

gives 56 as the result.

Which value of x is correct? Only the asker’s hairdresser knows for sure. GingerAle  Jun 22, 2017
#9
+2

You are right GingerAle - these multiple exponentiations are making my brain swirl!

Alan  Jun 22, 2017
#10
0

Yea they always confuse me, I can never remember the most basic conventions. :(

Melody  Jun 22, 2017
#11
0

I know the feeling, Sir Alan. I sometimes have to take Dramamine when doing some of the math problems on here. GingerAle  Jun 22, 2017
#12
0

Ah, yes....I see that Alan's interpretation is correct.....but, as he said.....these are sometimes difficult to unravel....!!!   Jun 22, 2017