Given that 3 to the power of -n= 0.2
find the value of (3 to the power of 4) to the power of n.
so 3^-n=0.2
find value of (3^4)^n
my working out:
1)3=n surd 0.2
2)n must be decimal number
3) cuberoot o.2=0.5848035476425733
4)so n=0.5848035476425733
5)(3^4)^0.5848035476425733=13.0643801669
6) so that answer is the value of (3^4)^n
3^-n = 0.2
3^n = 1/0.2 = 5
(3^4)^n = (3^n)^4 = 5^4 = 625
we don't trust you
it’s 625
not answer
To find the value of (3^4)^n, let's first solve the equation 3^(-n) = 0.2.
1) Start by taking the logarithm of both sides of the equation to eliminate the exponent. For this case, let's use the natural logarithm (ln).
ln(3^(-n)) = ln(0.2)
2) Apply the exponent rule: ln(a^b) = b * ln(a).
-n * ln(3) = ln(0.2)
3) Divide both sides of the equation by ln(3) to isolate "-n":
-n = ln(0.2) / ln(3)
4) Use a calculator to evaluate the right side of the equation:
-n = -1.609437912 / 1.098612289
-n ≈ -1.464973520
5) Multiply both sides by -1 to solve for "n":
n ≈ 1.464973520
Now that we have the value of "n", we can substitute it into the expression (3^4)^n:
(3^4)^n = 3^(4n)
Substituting the value of "n" we found earlier:
(3^4)^1.464973520 ≈ 13.0643801669
So, the value of (3^4)^n is approximately 13.0643801669.