How do you solve $10^{2 x} + 3 = 8$ $10^(2x)+3=8$ ?

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Answer 1 · 2016-07-09T23:38:55

$10^{2 x} + 3 = 8$ $10^(2x)+3=8$

$10^{2 x} = 5$ $10^(2x)=5$

$2 x \cdot log (10) = log (5)$ $2x * log(10) = log(5)$

$2 x = \frac{log (5)}{log (10)}$ $2x = log(5) / log(10)$

$x = (\frac{log (5)}{log (10)}) \cdot \frac{1}{2}$ $x = (log(5) / log(10)) * 1/2$

$x = \frac{log (5)}{2 log (10)}$ $x = log(5) / (2 log(10))$

Remember that ${log}_{b} (x^{r}) = r {log}_{b} (x)$ $log_b (x^r) = r log_b (x)$ .

How do you solve 102x+3=810^(2x)+3=8?