Question 354347
a) {{{e^(3x) >= 21}}}
Find the natural log (because of the "e" and because we want a decimal):
{{{ln(e^(3x)) >= 21}}}
Use a property of logarithms, {{{log(a, (p^q)) = q*log(a, (p))}}}, to move the exponent out in front:
{{{3x*ln(e) >= 21}}}
By definition, ln(e) = 1 so this simplifies to:
{{{3x >= 21}}}
Divide both sides by 3:
{{{x >= 21}}}<br>
b) {{{(1/5)^(t-2) = 125}}}
The quick way to solve this comes from recognizing that both sides are powers of 5: {{{1/5 = 5^(-1)}}} and {{{125 = 5^3}}}. Subsituting these in we get:
{{{(5^(-1))^(t-2)) = 5^3}}}
On the left side the rule is to multiply exponents:
{{{5^(-t+2) = 5^3}}}
Now the equation explicitly says each side is a power of 5. And if these two powers of 5 are equal, then the exponents must be equal!
{{{-t + 2 = 3}}}
Solving this:
{{{-t = 1}}}
{{{t = -1}}}