How do you solve #2log_2x-log_2 5=2^2#?

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Answer 1 · 2016-04-10T09:59:44

#x=4sqrt5#

As #log_ba=loga/logb#

#2log_2 x-log_2 5=2^2# can be written as

#2logx/log2-log5/log2=4# and

multiplying each side by #log2# as #log2!=0#, we get

#2logx-log5=4log2# or

#x^2/5=2^4# or #x^2=80# or

#x=sqrt80=4sqrt5# (as #x# cannot be negative)

Answer 2 · 2016-04-10T13:41:03

#color(blue)(=>x=4sqrt(5))#

Example: I have chosen to use log to base 10 so that you can check it on your calculator if you so wish.

suppose we had #Log_10(z)=2#

This means# -> 10^2=z#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the value of "x)#

Given: #" "2log_2x-log_2 5=2^2#

Write as: #log_2(x^2) -log_2( 5)=4#

Subtraction of logs means that the source value are applying division

#=> log_2(x^2/5)=4#

Using the principle demonstrated in my example

#=> log_2(x^2/5)=4" "->" "2^4=x^2/5#

#=> x=sqrt(80)" "=" " sqrt( 2^2xx2^2xx5)#

#color(blue)(=>x=4sqrt(5))#