Solve inequalities of the form a^x > or < b Algebra II

Solving a^x\leq b requires taking the log (with any base) of both sides. Using natural log (increasing function), the equation becomes:

\ln\left(a^x\right)\leq \ln\left(b\right)

x\cdot \ln\left(a\right)\leq \ln\left(b\right)

Therefore, the solution to 3^x\leq 31 is found by taking the natural log of both sides:

x\cdot \ln\left(3\right)\leq \ln\left(31\right)

Dividing both sides by \ln\left(3\right) (positive real number) gives:

x\leq\dfrac{\ln\left(31\right)}{\ln\left(3\right)}

Solving a^x \gt b requires taking the log (with any base) of both sides. Using natural log (increasing function), the equation becomes:

\ln\left(a^x\right) \gt \ln\left(b\right)

x\cdot \ln\left(a\right) \gt \ln\left(b\right)

Therefore, the solution to 5^x \gt 15 is found by taking the natural log of both sides:

x \cdot \ln\left(5\right) \gt \ln\left(15\right)

Dividing both sides by \ln\left(5\right) (positive real number) gives:

x \gt \dfrac{\ln\left(15\right)}{\ln\left(5\right)}

Solving a^x \gt b requires taking the log (with any base) of both sides. Using natural log (increasing function), the equation becomes:

\ln\left(a^x\right) \gt \ln\left(b\right)

x\cdot \ln\left(a\right) \gt \ln\left(b\right)

Therefore, the solution to 2 \cdot 4^x \gt 6 is found by taking the natural log of both sides:

2 \cdot 4^x \gt 6

4^x \gt 3

\ln\left(4^x\right) \gt \ln\left(3\right)

x \cdot \ln\left(4\right) \gt \ln\left(3\right)

Dividing both sides by \ln\left(4\right) (positive real number) gives:

x \gt \dfrac{\ln\left(3\right)}{\ln\left(4\right)}

Solving a^x\geq b requires taking the log (with any base) of both sides. Using natural log (increasing function), the equation becomes:

\ln\left(a^x\right)\geq \ln\left(b\right)

x\cdot \ln\left(a\right)\geq \ln\left(b\right)

Therefore, the solution to 3^{6x} \geq 42 is found by taking the natural log of both sides:

6x \cdot \ln\left(3\right) \geq \ln\left(42\right)

Dividing both sides by 6\ln\left(3\right) (positive real number) gives:

x \geq \dfrac{\ln\left(42\right)}{6\ln\left(3\right)}

Solving a^x \lt b requires taking the log (with any base) of both sides. Using natural log (increasing function), the equation becomes:

\ln\left(a^x\right) \lt \ln\left(b\right)

x\cdot \ln\left(a\right) \lt \ln\left(b\right)

Therefore, the solution to \left(7 \cdot 3^8\right)^x \lt 5 is found by taking the natural log of both sides:

\ln\left(7 \cdot 3^8\right)^x \lt \ln\left(5\right)

x \cdot \ln\left(7 \cdot 3^8\right) \lt \ln\left(5\right)

Dividing both sides by \ln\left(7 \cdot 3^8\right) (positive real number) gives:

x \lt \dfrac{\ln\left(5\right)}{\ln\left(7 \cdot 3^8\right)}

Solving a^x \gt b requires taking the log (with any base) of both sides. Using natural log (increasing function), the equation becomes:

\ln\left(a^x\right) \gt \ln\left(b\right)

x\cdot \ln\left(a\right) \gt \ln\left(b\right)

Therefore, the solution to 6^{\frac{-3}{2}x} \gt 3 is found by taking the natural log of both sides:

\ln\left(6^{\frac{-3}{2}x}\right) \gt \ln\left(3\right)

\left(\dfrac{-3}{2}x\right)\ln\left(6\right) \gt \ln\left(3\right)

Dividing both sides by \dfrac{-3}{2}\ln\left(6\right) (negative real number) gives:

x \lt \dfrac{\ln\left(3\right)}{\dfrac{-3}{2}\ln\left(6\right)}

x \lt \left(\dfrac{-2}{3}\right)\dfrac{\ln\left(3\right)}{\ln\left(6\right)}

Solving a^x\leq b requires taking the log (with any base) of both sides. Using natural log (increasing function), the equation becomes:

\ln\left(a^x\right)\leq \ln\left(b\right)

x\cdot \ln\left(a\right)\leq \ln\left(b\right)

Therefore, the solution to 8^x \leq 145 is found by taking the natural log of both sides:

\ln\left(8^x\right) \leq \ln\left(145\right)

x \cdot \ln\left(8\right) \leq \ln\left(145\right)

Dividing both sides by \ln\left(8\right) (positive real number) gives:

x \leq \dfrac{\ln\left(145\right)}{\ln\left(8\right)}