`(1/3)^x=(sqrt(3))^(x+2)` Solve the equation.Please show your workings clearly.Thanks
To solve, express the square root as an exponent.
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` .
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).
Then, divide both sides by log3 to simplify the equation.
Also, multiply both sides by 2.
Next, bring together the terms with x. So, subtract both sides by x.
And, divide both sides by -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
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` .