How do you solve $5^{x + 2} = 4$ $5^(x+2)=4$ ?

Question

12611 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-25T15:11:46

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

One of the logarithm rules one should keep in mind for this problem:

${log a}^{b} = b \cdot log a$ $log a^b = b*loga$

Apply logarithm on both sides

$log (5^{x + 2}) = log 4$ $log(5^(x+2)) = log 4$

$\Rightarrow (x + 2) \cdot log 5 = log 4$ $=>(x+2)*log 5 = log 4$

$\Rightarrow x + 2 = \frac{log 4}{log 5}$ $=>x+2 = log 4/log 5$

Now it's just a matter of simplification:

$\Rightarrow x = \frac{log (2^{2})}{log 5} - 2$ $=> x = log(2^2)/log 5 - 2$

$\Rightarrow x = \frac{2 \cdot log 2}{log 5} - 2$ $=>x = (2*log 2)/ log 5 - 2$

$\Rightarrow x = \frac{2 \cdot log 2 - 2 log 5}{log 5}$ $=> x = (2*log 2 - 2 log 5)/log 5$

$or, x = \frac{2 \cdot (log 2 - log 5)}{log 5}$ $or, x = (2*(log 2 - log 5))/log 5$

How do you solve 5x+2=45^(x+2)=4?