Question #d4055

2 Answers
Mar 6, 2016

You need to clarify the question
To format use: hash symbol log_5(2x+1) hash symbol
This looks like: #Log_5(2x+1)#

Explanation:

Assumption: The question is:

#log_3(x)=-2# ......................................(1)
#log_5(2x+1)-log_5(x)=2#................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Pre amble:")#

Compare to #" "log_10(x)=3#
Another way of writing this is: #10^3=x#

Also #log_z(a) -log_z(b) = log_z(a/b)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Consider equation (1)")#

Write as #" "3^(-2)=x#

#color(blue)(=>x-1/3^2=1/9) ..........................(1_a)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Consider equation (2)")#

Write as:#" "log_5((2x+1)/2)=2#

#=>5^2=(2x+1)/2#

#=>25=x+1/2#

#color(blue)(=>x=24.5)..............................(2_a)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(red)("Either I have interpreted the questions incorrectly or")#
#color(red)("the two equation are not connected!")#
#color(red)("It could be that the question is wrong!")#

#color(magenta)("Equation "(2_a) !=" Equation "(1_a)" "->" contradiction")#

Mar 6, 2016

A different interpretation of question

Explanation:

Assumption:

#log(3^x)=-2#................................(1)
#log(5^(2x+1))-log(5^x)=2#.... ..(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Consider equation (1)

Write as #" "xlog(3)=2#

#color(blue)(=>x=2/log(3) ~~4.1918)# to 4 decimal places

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider equation (2)

Write as #" "(2x-1)log(5)-xLog(5)=2#

#=>log(5)(2x-1-x)=2#

#x-1=2/log(5)#

#color(blue)(x= 2/log(5)+1 ~~3.8613)# to 4 decimal places