How do you solve $3^{2 x + 1} = 5$ $3^(2x+1) = 5$ ?

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Answer 1 · 2017-08-03T16:33:09

$x = 0.232$ $x = 0.232$ (to 3 sf)

We will not attempt to treat this as an exponential equation by working with the bases, because $5$ $5$ is not a power of $3$ $3$ .

Find the log of both sides.

${log 3}^{2 x + 1} = log 5 \leftarrow$ $log 3^(2x+1) = log5" "larr$ use the log power law

$(2 x + 1) log 3 = log 5 \leftarrow$ $(2x+1) log3 = log5" "larr$ isolate the $x$ $x$ term

$2 x + 1 = \frac{log 5}{log 3}$ $2x+1= log5/log3$

$2 x = \frac{log 5}{log 3} - 1$ $2x = log5/log3-1$

$x = (\frac{log 5}{log 3} - 1) \div 2$ $x = (log5/log3-1)div2$

There are no laws to simplify this, so use a calculator to find a value for $x$ $x$

$x = 0.23248676$ $x = 0.23248676$

How do you solve 32x+1=53^(2x+1) = 5?