color(blue)(2)log_3 (4x-5)= log_3(4x+5) + log_3(4x-3)2log3(4x−5)=log3(4x+5)+log3(4x−3)
color(white)(xxxxxxxxxxxxxxxxxx.x)darr×××××××××.x↓
log_3 (4x-5)^color(blue)(2)= log_3[color(red)((4x+5) xx (4x-3))]log3(4x−5)2=log3[(4x+5)×(4x−3)]
We have used using 2 log laws:
color(blue)(log_a b^c hArr clog_a b) and color(red)(log_a b + log_a c = log_a(bxxc))logabc⇔clogabandlogab+logac=loga(b×c)
Now we can write:
(4x-5)^2= (4x+5) xx (4x-3)(4x−5)2=(4x+5)×(4x−3)
This uses another log law: log_a B = log_a C hArr B=ClogaB=logaC⇔B=C
Now solve it like a normal equation .... remove the brackets:
16x^2-40x+25 = 16x^2-12x+20x-1516x2−40x+25=16x2−12x+20x−15
1cancel(6x^2)-40x+25 = cancel(16x^2)-12x+20x-15
25+15 = 8x+40x" "larr re-arrange, keep x positive.
40 = 48x
40/48 = x
5/6 = x
Notice that, (4(5/6)-5) < 0" "larr log impossible
However, the fact that it is squared gives a positive value, and the log value can then be determined.
