Question

How do you solve `3^ { 2\log x } + 9= 10\cdot 3\log x`?

Answer 1

`x = {1, e^2}`

Explanation:

Assuming that the equation reads

`3^(2logx)+9=10 * 3^(log x)` and now making `3^(log x)= y` we have

`y^2-10 y +9=0->{(y_1=9),(y_2=1):}`

then

`3^(log x) = 9 -> logx log 3=2 log 3->log x = 2->x = e^2`
`3^(log x)=1->log x=0->x = 1`