Question

Question #4c938

Answer 1

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`(2/3)^(-x)+4=8/27`

`(2/3)^(-x)=8/27-4=(8-4*27)/27=(8-108)/27=-100/27`

Now, we can take logs of both sides:

`Log((2/3)^(-x))=Log(-100/27)`

Log of a negative number is undefined.

There is no solution for this in real numbers.

To solve it you would have to deal with complex numbers.

Answer 2

`x=1`

Explanation:

.

`(2/3)^(-x+4)=8/27`

Whenever you have your variable in the exponent you need to use logarithms to solve the equation:

We take logs of both sides. It makes no difference whether you use common logs or natural logs. Whichever you use, you should use the same type on both sides. We will use natural logs here:

`Ln((2/3)^(-x+4))=ln(8/27)`

Now, we use the law of logarithms that says the exponent of the term you are taking log of can move behind the log and become a coefficient:

`(-x+4)ln(2/3)=ln(8/27)`

There is another log rule that says:

`ln(a/b)=lna-lnb`. Using this rule:

`(-x+4)(ln2-ln3)=ln8-ln27`

`(-x+4)(ln2-ln3)=ln(2^3)-ln(3^3)`

`(-x+4)(ln2-ln3)=3ln2-3ln3`

`(-x+4)(ln2-ln3)=3(ln2-ln3)`

Now, we divide both sides by `(ln2-ln3)`:

`((-x+4)(ln2-ln3))/(ln2-ln3)=(3(ln2-ln3))/(ln2-ln3)`

`((-x+4)cancelcolor(red)(ln2-ln3))/cancelcolor(red)(ln2-ln3)=(3cancelcolor(red)(ln2-ln3))/cancelcolor(red)(ln2-ln3)`

`-x+4=3`

`-x=-1`

`x=1`