What is the answer to (\sqrt[5]{7})^x=7^{2x+3} ?

The answer to (\sqrt[5]{7})^x=7^{2x+3} is x=-5/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\sqrt[5]{7})^x=7^{2x+3}

x = - \frac{5}{3}

Decimal Notation

x = - 1.66666 \dots

Solution steps

Apply exponent rules

\frac{1}{5} x = 2 x + 3

Solve

\frac{1}{5} x = 2 x + 3 : x = - \frac{5}{3}

x = - \frac{5}{3}

e^{3 x - 5} = 13

3 2^{p} = 2

3^{3 b} = 243

2^{x + 2} = 8^{x - 2}

e^{2 x} = \frac{1}{32}

What is the answer to (\sqrt[5]{7})^x=7^{2x+3} ?
The answer to (\sqrt[5]{7})^x=7^{2x+3} is x=-5/3
Solve for x: (\sqrt[5]{7})^x=7^{2x+3}
The solution is x=-5/3