Mennyire kell hozzá bő magyarázat? Gondolom a fogalmakkal tisztában vagy, ha nem, akkor át kell nézni.

a,

`2^(x+3)+2^(x+1)=10`

`2^(x+3)` = `2^x*2^3` = `2^x*8`

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

`2^(x+3)+2^(x+1)` = `2^x*8+2^x*2` = `2^x*(8+2)` = `2^x*10`

`2^x*10` = 10 /:10

`2^x` = 1

Hányadik hatvány lesz 1? nulladik.

`2^0` = 1

x = 0

Ellenőrzés:

`2^(0+3)+2^(0+1)=10`

`2^3+2^1=10`

`8+2=10`

10 = 10

b,

`5^(x+4)-5^(x+2)=24`

`5^(x+4)` = `5^x*5^4` = `625*5^x`

`5^(x+2)` = `5^x*5^2` = `25*5^x`

`5^(x+4)-5^(x+2)` = `625*5^x-25*5^x` = `5^x*(625-25)` = `600*5^x`

`600*5^x=24` /:600

`5^x=24/600` = `(cancel(2^3)*cancel(3))/(cancel(2^3)*cancel(3)*5^2)` = `1/5^2` = `5^(-2)`

Ha az alapok megegyeznek, a kitevők is egyenlők.

x = -2

c,

`2*3^(x-1)+3^(x+3)=83`

`2*3^(x-1)` = `2*3^x*3^(-1)` = `2*3^x*1/3` = `2/3*3^x`

`3^(x+3)` = `3^x*3^3` = `3^x*27`

`2*3^(x-1)+3^(x+3)` = `2/3*3^x+3^x*27` = `3^x*(2/3+27)` = `27 2/3*3^x`

`27 2/3 *3^x = 83` /:`27 2/3`

`3^x` = `83/(27 2/3)` = 3 = `3^1`

x = 1

d,

`3^x-3^(x-1)-7*3^(x-3)=33`

`3^(x-1)` = `3^x*3^(-1)` = `3^x*1/3`

`7*3^(x-3)` = `7*3^x*3^(-3)` = `7*3^x*1/3^3` = `7*3^x*1/27` = `7/27*3^x`

`3^x-3^(x-1)-7*3^(x-3)` = `3^x-3^x*1/3-7/27*3^x` = `3^x*(1-1/3-7/27)` = `3^x*((27-9-7)/27)` = `3^x*11/27`

`3^x*11/27 = 33` /:`11/27`

`3^x` = `33/(11/27)` = `33*(27/11)` = `(3*3^3*cancel(11))/cancel(11)` = `3^1*3^3` = `3^(1+3)` = `3^(4)`

x = 4.