5.02 The exponential function

Rewrite each side of the equation with a base of 2.

Hence, solve for x.

8^{x} = 4

16^{x} = \dfrac{1}{2}

\dfrac{1}{1024} = 4^x

\left(\sqrt{2}\right)^{x} = \sqrt[5]{32}

4^{x} = 4^{8}

3^{x} = 3^{\frac{2}{9}}

3^{x} = 27

7^{x} = 1

8^{x} = \dfrac{1}{8^{2}}

3^{y} = \dfrac{1}{27}

10^{x} = 0.01

3^{x} = 3^{6}

6^{x} = 6^{ - 3 }

6^{x} = 6^{\frac{4}{3}}

2^{x} = 64

9^{x} = 1

5^{x} = \dfrac{1}{5^{2}}

10^{x} = 0.0001

9^{y} = 81

9^{y} = 27

25^{y} = 125

3^{ 5 x - 10} = 1

25^{x + 1} = 125^{ 3 x - 4}

9^{x + 4} = 27^{x}

3^{ 4 x - 8} = 1

8^{x + 3} = 32^{ 2 x - 1}

30 \times 2^{x - 6} = 15

2^{x} \times 2^{x + 3} = 32

3^{x} \times 9^{x - k} = 27

a^{x + 1} = a^{3} \sqrt{a}

3^{x^{2} - 3 x} = 81

27 \left(2^{x}\right) = 6^{x}

24 \times 2^{x - 6} = 12

2^{x} \times 2^{x + 2} = 16

3^{x} \times 9^{x - k} = 9

5^{x} = \sqrt[3]{5}

30^{n} = \sqrt[3]{30}

5^{x} = \sqrt{5}

5^{x} = \sqrt[4]{5}

9^{x} = \sqrt[9]{9}

\left(\sqrt{7}\right)^{y} = 49

\left(\sqrt{6}\right)^{y} = 36

\left(\dfrac{5}{8}\right)^{x} = \left(\dfrac{5}{8}\right)^{9}

7 \left(4^{x}\right) = \dfrac{7}{4^{3}}

\dfrac{1}{3^{x - 3}} = \sqrt[3]{9}

\left(\dfrac{1}{9}\right)^{x + 5} = 81

\left(\dfrac{1}{8}\right)^{x - 3} = 16^{ 4 x - 3}

\dfrac{25^{y}}{5^{4 - y}} = \sqrt{125}

8^{x + 5} = \dfrac{1}{32 \sqrt{2}}

\left(\dfrac{1}{25}\right)^{x - 4} = 125^{ 3 x - 1}

8^{x + 4} = \dfrac{1}{32 \sqrt{2}}

\left(\dfrac{1}{27}\right)^{x - 5} = 81

\dfrac{1}{5^{x + 3}} = \sqrt[3]{25}

9 \left(8^{x}\right) = \dfrac{9}{8^{2}}