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

Question

1925 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-09T10:37:51

This equation has one solution #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^(2x+1)-2=0# or #5^(x+1)+5=0#

#5^(2x+1)=2# or #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^(2x+1)=5^(log_5 2)#

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

#2x+1=log_5 2#
#2x=log_5 2-1#

#x=(log_5 2-1)/2#