Solve the exponential equation algebraically
5^(-t/2) = 0.20
5^(-t/2)=0.20
take log on both sides,
(-t/2)log(5)=log(0.20)
solve for t
-t/2 = log(0.20)/log(5)
t=-2log(0.20)/log(5)...isolate t
=log(1/0.04)/log(5)...absorb -2 into log(0.2)
=log(25)/log(5)... evaluate reciprocal
=2... laws of logarithm, 25=5²
To solve the exponential equation algebraically, we need to isolate the variable on one side of the equation.
Starting with the given equation:
5^(-t/2) = 0.20
We can rewrite 0.20 as a fraction:
5^(-t/2) = 1/5
To remove the exponent, we can take the logarithm of both sides of the equation. In this case, we can use the natural logarithm (ln) or the logarithm to any base:
ln(5^(-t/2)) = ln(1/5)
Now, we can apply the power rule for logarithms which states that ln(a^b) = b * ln(a):
(-t/2) * ln(5) = ln(1/5)
Next, we can simplify the equation by using the logarithmic property that ln(1/x) = -ln(x):
(-t/2) * ln(5) = -ln(5)
We can divide both sides of the equation by ln(5) to isolate the variable:
(-t/2) = -ln(5) / ln(5)
Simplifying further gives:
(-t/2) = -1
To solve for t, we can multiply both sides of the equation by 2:
t = 2
Therefore, the solution to the exponential equation 5^(-t/2) = 0.20 is t = 2.