those logs are just numbers, so
x log19 - 5log19 = x log3 + 2log3
x(log19-log3) = 2log3 + 5log19
x =
2log3 + 5log19
--------------------
log19 - log3
Solve: 19^(x-5)=3^(x+2)
i know you then go: (x-5) log19= (x+2) log3
i just don't know what goes after?
2 answers
expand:
(x-5) log19= (x+2) log3
xlog19 - 5log19 = xlog3 + 2log3
xlog19 - xlog3 = 2log3 + 5log19
x(log19 - log3) = 2log3 + 5log10
x = (2log3 + 5log19)/(log19 - log3)
= appr 9.1663
(x-5) log19= (x+2) log3
xlog19 - 5log19 = xlog3 + 2log3
xlog19 - xlog3 = 2log3 + 5log19
x(log19 - log3) = 2log3 + 5log10
x = (2log3 + 5log19)/(log19 - log3)
= appr 9.1663