Question

30=50(1-0.9^x)

Answer 1

are u solving for x?

Answer 2

yes

Answer 3

30= (50 X 1 X -0.9^x)

Answer 4

i tried to solve it but you can't take the log of a negative to get x

Answer 5

30= 50 - 45^x

Answer 6

why would you distribute the 50

Answer 7

Hmmm....
30=50(1-0.9^x)
3/5=1-0.9^x
0.9^x=2/5

Then switch to log form:

\[\Large \log_{0.9} \frac{ 2 }{ 5 }=x\]

You'll need the change of base formula to get a numerical result.

Answer 8

okay. that works. Makes sense! thanks!

Answer 9

You can't distribute the 50 like that. PEMDAS... exponents before multiplication.

At best, you can:

30=50(1-0.9^x)
30=50-50(0.9^x)

But that doesn't really help matters. :)

Answer 10

You're welcome. :)

Answer 11

you don't need change of base... apply the log law for powers

xln(0.9) = ln(2/5)

solve for x

Answer 12

or you could use a calculator and do log(2/5)/log(0.9)

Answer 13

Yup, that works too.... just depends on what methods you've learned and prefer. It's really the same thing. :)

Answer 14

@fullib , that is the change of base formula. :) You can either use Log or Ln.

Answer 15

Notice that from:
xln(0.9) = ln(2/5)

You get: x=ln(2/5) / ln(0.9)

Change of base says that;
\[\Large \log_{0.9} \frac{ 2 }{ 5 }=\frac{ \ln(2/5) }{ \ln0.9 }\]

So they are really just two slightly different routes to the exact same result. :)

Answer 16

so make the solution easier... and minimising mistake by eliminating unnecessary steps