How do you solve #\log _ { 5} ( x + 52) + \log _ { 5} x = \log _ { 5} 53#?

1 Answer
George C.
Apr 19, 2017

#x=1#

Explanation:

Given:

#log_5(x+52) + log_5(x) = log_5(53)#

#color(white)()#
Method 1

Note that for any valid base #b#, we have:

#log_b(1) = 0#

So:

#log_5(color(blue)(1)+52) + log_5(color(blue)(1)) = log_5(53) + 0 = log_5(53)#

So #x=1# is a solution.

#color(white)()#
Method 2

If #b# is a valid base and #c, d > 0# then:

#log_b c + log_b d = log_b cd#

So, provided #x > 0#, we find:

#log_5(53) = log_5(x+52) + log_5(x)#

#color(white)(log_5(53)) = log_5((x+52)x)#

#color(white)(log_5(53)) = log_5(x^2+52x)#

Note that #log_b# is one to one as a real valued function, so we must have:

#53 = x^2+52x#

Subtract #53# from both sides to get:

#0 = x^2+52x-53 = (x-1)(x+53)#

So #x=1# or #x=-53#

But considered as a real valued function, there is no value of #y# such that #5^y = -53#. So #log_5(-53)# is not real valued.

So the only valid real solution is #x=1#

#color(white)()#
Complex footnote

It is possible to define #log_5(x)# for negative values of #x#.

The principal value is #log_5(-x)+pi/ln 5 i#

Note that with this definition, #x=-53# is still not a solution of the given equation:

#log_5(color(blue)(-53)+52) + log_5(color(blue)(-53))#

#= log_5(-1)+log_5(-53)#

#= (log_5(1)+pi/ln 5i)+(log_5(53)+pi/ln5i)#

#= log_5(53)+(2pi)/ln5i#

#!= log_5(53)#

Related questions
Impact of this question
644 views around the world
You can reuse this answer
Creative Commons License