Question

Aflati-l pe x din egalitatile:
a)2^x+3=128
b)3^2x-1=243
c)4^2x+3=64
d)4^x+4^x+1+4^x+2=336
e)2^2x+3+4^x+1=896
f)9^6x+81^3x+27^4x=245^5

Answer 1

a) 2la x+ 3=128
2la x+3=2 la 7
[baza este aceeasi=2  ,deci egalam exponentii]
x+3=7
x=4

b)3la 2x-1=243
3la 2x-1=3la 5
[avem aceeasi baza=3,egalam exponentii]
2x-1=5
2x=6
x=3

c)4la 2x+3=64
4la 2x+3=4la 3
2x+3=3
2x=0
x=0

d)4lax +4la x+1+4la x+2=336
[dam factor comun pe 4 la x]
4la x(1+4+4la2)=336
4lax(5+16)=336
4la x  X21=336
4la x=336:21
4lax=16
4lax =4la 2
x=2

f)9la 6x+81la 3x+27la 4x=245la 5
(3la2)la6x+(3la4)la3x+(3la3)la4x=245 la5  [verifica eu cred ca este 243 la 5]
                                                                         [243=3la puterea 5]
3la12x+3la 12x+3la12x=(3la 5)la 5
3 ori 3la12x=3la25
3la (12x+1)=3la 25
12x+1=25
12x=24
x=24:12
x=2

Answer 2

a)2^x+3=128
2^x·2³=2⁷
2^x=2⁷:2³
2^x=2⁷⁻³
2^x=2⁴
x=4

b)3^2x-1=243
3^2x·3⁻¹=3⁵
3^2x=3⁵
3^2x·1/3=3⁵
3^2x=3⁵:1/3
3^2x=3⁵·3
3^2x=3⁵⁺¹
3^2x=3⁶
2x=6
x=6/2
x=3

c)4^2x+3=64
4^2x·4³=4³
4^2x=4³:4³
4^2x=4³⁻³
4^2x=4⁰
2x=0
x=0

d)4^x+4^x+1+4^x+2=336
4^x+4^x·4+4^x·4²=336
4^x·(1+4+16)=336
4^x·21=336
4^x=336:21
4^x=16
4^x=4²
x=2