# Is it possible to solve this equation without using a calculator, no imaginairy numbers? and how :) 9 * (3^x) = 4 * (2^x) find x x = -2 the...

Is it possible to solve this equation without using a calculator, no imaginairy numbers? and how :)

**9 * (3^x) = 4 * (2^x) find x x = -2**

the way i tried:

9 * (3^x) = 4 (2^x)

(3²) * (3^x) = (2²) * (2^x)

3^(x+2) = 2^(x+2)

log(3^(x+2)) = log(2^(x+2))

(x+2) * log (3) = (x+2) * log (2)

xlog(3) + 2log(3) = xlog(2) + 2log(2)

xlog(3) - xlog(2) = 2log(2) - 2log(3)

x * (log(3) - log(2)) = 2log(2) - 2log(3)

x = (2log(2) - 2log(3)) / (log(3) - log(2)) --> next step?

x = -2 (with calculator)

*print*Print*list*Cite

Finishing your method:

`9*3^x=4*2^x`

`3^(x+2)=2^(x+2)`

`(x+2)log3=(x+2)log2`

`x(log3-log2)=2log2-2log3`

`x=(2log2-2log3)/(log3-log2)`

`=(-2(log3-log2))/(log3-log2)` Factor -2 out of both terms in the numerator

`=-2`

You did excellent work getting there! (Why factor -2? It makes the expressions in the numerator and the denominator the same)

---------------------------------------------------------------

Alternatively you can stop here:

(x+2)log3=(x+2)log2

Since `log3!=log2` , but x+2=x+2, the only way for this equation to be true is if x+2=0 ==> x=-2. (If ab=ac and `b!=c` then a=0 -- this is a corollary to the multiplication property of equality that alklows you to multiply both sides of an equation by any nonzero number and retain equality.)