How do you solve #log_x7 = 1#?

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Answer 1 · 2015-10-26T20:11:01

#x = 7#

#log_a(b) =ln(b)/ln(a)#

So

#log_x(7) = ln(7)/ln(x)#

#=> ln(7)/ln(x) = 1#

#=>ln(7) = ln(x)#

Taking exponentiel both side

#=>x = 7#

Answer 2 · 2015-10-26T20:20:08

x=7

Consider powers of 10
Picking one at random

#10^2 = 100#

If this were to be written as log base 10 it would be:

# log_10(100) = 2#

Following the same approach for your question but in you case we could reverse the process to get something we can work out.

so #log_x(7) = 1 " "->" " x^(1) = 7 #

Anything raised to the power of one is its own value

So #x^1 = x = 7#

This means that if z is any number (technically you would have to say that #z in R# but I would not wary about that!)

Then #log_z(z) = 1#