How do you solve $[(5^{2 x + 1}) - 2] \cdot [(5^{x + 1}) + 5] = 0$ $[(5^(2x+1)) - 2] * [(5^(x+1)) + 5] = 0$ ?

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Answer 1 · 2015-08-09T10:37:51

This equation has one solution $x = \frac{{log}_{5} 2 - 1}{2}$ $x=(log_5 2-1)/2$

On left side there is a multiplication of 2 expressions, so it is zero, when any of those expressions is:

$5^{2 x + 1} - 2 = 0$ $5^(2x+1)-2=0$ or $5^{x + 1} + 5 = 0$ $5^(x+1)+5=0$

$5^{2 x + 1} = 2$ $5^(2x+1)=2$ or $5^{x + 1} = - 5$ $5^(x+1)=-5$

The second equation has no solutions, because exponential functions do not have negative values.

To solve the first equation we have to change 2 to a power of 5.

$5^{2 x + 1} = 5^{{log}_{5} 2}$ $5^(2x+1)=5^(log_5 2)$

Now, when we have the same base on both sides we can skip it to get

$2 x + 1 = {log}_{5} 2$ $2x+1=log_5 2$
$2 x = {log}_{5} 2 - 1$ $2x=log_5 2-1$

$x = \frac{{log}_{5} 2 - 1}{2}$ $x=(log_5 2-1)/2$

How do you solve [(52x+1)−2]⋅[(5x+1)+5]=0 [(5^(2x+1)) - 2] * [(5^(x+1)) + 5] = 0?