# How do you solve  3^(2x+1) = 5?

Aug 8, 2016

$x = 0.2326$.

#### Explanation:

${3}^{2 x + 1} = 5$

$\therefore {3}^{2 x} \cdot 3 = 5$

$\therefore {3}^{2 x} = \frac{5}{3}$

$\therefore {\log}_{10} {3}^{2 x} = {\log}_{10} \left(\frac{5}{3}\right)$

$\therefore 2 x \cdot {\log}_{10} 3 = {\log}_{10} 5 - {\log}_{10} 3$

$\therefore 2 x \left(0.4771\right) = 0.6990 - 0.4771$................[Using, Log Tables].

$\therefore 2 x \left(0.4771\right) = 0.2219$

$\therefore {\log}_{10} \left\{\left(2 x\right) \left(0.4771\right)\right\} = {\log}_{10} 0.2219$

$\therefore {\log}_{10} 2 + {\log}_{10} x + {\log}_{10} 0.4771 = {\log}_{10} 0.2219$

$\therefore 0.3010 + {\log}_{10} x + \left(\overline{1}\right) .6786 = \left(\overline{1}\right) .3461 .$

$\therefore {\log}_{10} x + \left(\overline{1}\right) .9796 = \left(\overline{1}\right) .3461$

$\therefore {\log}_{10} x = 0.3461 - 0.9796 = - 0.6335 = \left(\overline{1}\right) .3665$

$\therefore x$=Anti-${\log}_{10} \left(\overline{1}\right) .3665$

$\therefore x = 0.2326$.

Aug 11, 2016

$x = 0.232487$

#### Explanation:

We have an exponential equation, but 5 is not one of the powers of 3, so logs are indicated, especially if the variable is in the index.

${3}^{2 x + 1} = 5$

$\log {3}^{2 x + 1} = \log 5$

$\left(2 x + 1\right) \log 3 = \log 5$

$2 x + 1 = \log \frac{5}{\log}$

There is no law for simplifying #(log)/log), use a calculator or tables.

$2 x + 1 = 1.4649735$

$2 x = 0.464974$

$x = 0.232487$