# `(1/3)^x=(sqrt(3))^(x+2)` Solve the equation.Please show your workings clearly.Thanks

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`(1/3)^x=sqrt3^(x+2)`

To solve, express the square root as an exponent.

`(1/3)^x=3^(1/2(x+2))`

To remove the x from the exponent, take the logarithm of both sides.

`log (1/3)^x = log3^(1/23(x+2))`

Then, apply the property `log a^m=m log a` .

`x log(1/3)=1/2(x+2)log3`

At the left side, apply the quotient property of logarithm which is `log(M/N)=logM-logN` .

`x (log1 - log3) = 1/2(x+2)log3`

Note that when the argument of the logarithm is 1, it is equal to zero (log1=0).

`x(0-log3)=1/2(x+2)log3`

`x(-log3)=1/2(x+2)log3`

Then, divide both sides by log3 to simplify the equation.

`(x(-log3))/log3=(1/2(x+2)log3)/(log3)`

`-x=1/2(x+2)`

Also, multiply both sides by 2.

`-x*2=(1/2(x+2))*2`

`-2x=x+2`

Next, bring together the terms with x. So, subtract both sides by x.

`-2x-x=x-x+2`

`-3x=2`

And, divide both sides by -3.

`(-3x)/(-3)=2/(-3)`

`x=-2/3`

**Hence, the solution to the given exponential equation is `x=-2/3` .**

To solve this in the easiest way:

`(1/3)^x = (sqrt3)^(x+2)` becomes

`3^-x = (3^(1/2))^(x+2)` using the rules of exponents. (Remember that `1/3` means `1 divide 3` and in terms of the rules of exponents when the expression is divided we subtract (-) the exponents.Thus

`3^(-x) = 3^(1/2 x + 2/2)` (having multiplied the exponents to remove the backet)

`therefore 3^-x = 3^(1/2x+1)` As the bases are the same so the exponents equal each other. Create a new equation:

`therefore -x = 1/2x +1`

`therefore -1 = 1/2x + x`

`therefore -1 = 3/2 x`

`therefore (-1 times 2)/3 = x`

`therefore x = -2/3`

**Ans: x = -2/3**

``

`sqrt(3)^(x+2)=(3^(1/2))^(x+2)=[(1/3)^(-1/2)]^(x+2)=(1/3)^(-x/2-1).`

For this to equal `(1/3)^x,` the exponents must be equal. Therefore we have `x=-x/2-1,` or `(3x)/2=-1,` **so **`x=-2/3` .