How do you solve #8^ { x + 1} = 19#?

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Answer 1 · 2017-03-03T02:18:07

#x=log_8 19 - 1#.

Starting with:

#8^(x+1)=19#

Take the log (to base 8) of both sides:

#log_8(8^(x+1))=log_8 19#

Because #log_8( * )# and #8^((*))# are inverse functions, they "cancel off":

#x+1=log_8 19#

Then, just subtract 1 from both sides:

#x+1-1=log_8 19 - 1#

#x=log_8 19 - 1#