Question

Resolva as equações exponenciais abaixo reduzindo a base

a) 2^3x=64
b) 3^5x+34=81
c) 5^x+1= 125
d) 81^6+x= 9 ^-2x

Answer 1

Resposta:

$2^3x=64\\\\\frac{2^3x}{8}=\frac{64}{8}\\\\x=8$

$3^5x+34=81\\\\243x+34=81\\\\243x+34-34=81-34\\\\243x=47\\\\\frac{243x}{243}=\frac{47}{243}\\\\x=\frac{47}{243}$

$5^x+1= 125\\\\5^x+1-1=125-1\\\\5^x=124\\\\x\ln \left(5\right)=\ln \left(124\right)\\\\x=\frac{\ln \left(124\right)}{\ln \left(5\right)}$

$81^6+x=9^{-2x}\\\\e^{4\ln \left(3\right)\left(\frac{u}{4\ln \left(3\right)}-81^6\right)}\left(81^6+\left(\frac{u}{4\ln \left(3\right)}-81^6\right)\right)=1\\\\\frac{e^{u-81^6\cdot \:4\ln \left(3\right)}u}{4\ln \left(3\right)}=1\\\\u=\text{W}\left(4\ln \left(3\right)\cdot \:81^{81^6}\right)\\\\x=\frac{\text{W}\left(4\ln \left(3\right)\cdot \:81^{81^6}\right)}{4\ln \left(3\right)}-8$

Answer 2

Explicação passo a passo:

a)

$2^{3x} = 64\\2^{3x} = 2^{6}\\3x = 6\\x= 2$

b)

$3^{5x} + 34 = 81\\3^{5x} = 47\\log3 (3^{5x}) = log3 (47)\\5x = log3(47)\\x = \frac{1}{5} .log3(47)$

c)

$5^{x} + 1 = 125\\5^{x} = 124\\log5(5^{x}) = log5(124)\\x = log5(124)$

d)

$81^{6+x} = 9^{-2x}\\(9^{2})^{6+x} = 9^{-2x}\\9^{12+2x} = 9^{-2x}\\12 + 2x = -2x\\4x = 12\\x = 3$