b,

I. `4*7^x+2^y=20`

II. `-3*7^x+2^y=61`

`7^x=a` és `2^y=b` helyettesítéssel

I. 4a+b=20

II. -3a+4b=61

I. *4: 16a+4b=80

II. - I.:

19a = 19

a = 1

I. `4*1+b=20`

b = 16

`7^x=1` `rightarrow` x = 0

`2^y=16` `rightarrow` y = 4

Ellenőrzés!

c,

I. `5^(2x)*5^y=5^5`

II. `6^(x-y)` = `6^1`

Ha az alapok azonosak, a kitevők is.

I. 2x+y = 5

II. x-y = 1

I. + II.:

3x = 6

x = 2

II.: 2-y = 1

y = 1

Ellenőrzés!


f,

I. `16^(2x)+16^(2y)= 36`

II. `16^(x+y)=8*root()(2)`

II. `2^(4(x+y))=2^(3+1/2)`

4x+4y = `7/2`

x+y = `7/8`

y = `7/8-x`

I. `16^(2x)+16^(7/4-2x)=36`

`256^x+128/256^x` = 36

`256^x=a`

`a^2+128=36a`

`a^2-36a+128=0`

`a_(1,2)` = `(36 pm root()(36^2-4*128))/2` = `(36 pm 28)/2`

`a_1` = 4 = `256^x`

`2^2=2^(8x)`

`8x=2`

`x_1` = `1/4` `rightarrow` `y_1` = `7/8-1/4` = `5/8`

`a_2` = 32 = `256^x`

`2^5=2^(8x)`

8x = 5

`x_2` = `5/8` `rightarrow` `y_2` = `7/8-5/8` = `2/8` = `1/4`

Megoldás:

x = `1/4`; y = `5/8`

x = `5/8`; y = `1/4`.