Question

calcule:

(3x-6)
log2= log5^(1-x)
obs: log2: 0,30 ​

Answer 1

$(3x-6).log2~=~log5^{(1-x)}\\\\\\(3x-6).log2~=~(1-x).log5\\\\\\(3x-6).log2~=~(1-x).log\left(\frac{10}{2}\right)\\\\\\(3x-6).log2~=~(1-x).\left(log10-log2\right)\\\\\\(3x-6).(0,30)~=~(1-x).(1-0,30)\\\\\\(3x-6).\frac{3}{10}~=~(1-x).\frac{7}{10}\\\\\\(3x-6)~.~3~=~(1-x)~.~7\\\\\\9x-18~=~7-7x\\\\\\9x+7x~=~7+18\\\\\\16x~=~25\\\\\\\boxed{x~=~\frac{25}{16}}$