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zlopas [31]
3 years ago
5

Can someone please show how to solve this ln(5x-1)+ln(x-6)=ln6 ???

Mathematics
2 answers:
Virty [35]3 years ago
8 0
When you add ln, you multiply
ln((5x-1)(x-6)=ln6  do now FOIL
ln(5x^2-30x-x+6)=ln6
therefore
5x^2-31x+6=6 and 
5x^2-31x=0 furthermore 
x(5x-31)=0
one solution is 0, which we will reject
the other solution is 5x=31
x=31/5
miv72 [106K]3 years ago
5 0
Ln(5x - 1) + ln(x - 6) = ln(6)
ln((5x - 1)(x - 6)) = ln(6)
            ln(5x(x - 6) - 1(x - 6) = ln(6)
ln(5x(x) - 5x(6) - 1(x) - 1(-6)) = ln(6)
ln(5x² - 30x - x + 6) = lm(6)
ln(5x² - 31x + 6) = ln(6)
     5x² - 31x + 6 = 6
5x² - 31x + 6 - 6 = 6 - 6
     5x² - 31x + 0 = 0
x = <u>-(-31) +/- √((-31)² - 4(5)(0))</u>
                       2(5)
x = <u>31 +/- √(961 - 0)</u>
                 10
x = <u>31 +/- √(961)
</u>              10<u>
</u>x = <u>31 +/- 31
</u>           10<u>
</u>x = <u>31 + 31</u>    x = <u>31 - 31
</u>          10                  10<u>
</u>x = <u>62</u>             x = <u>0
</u>      10                  10<u>
</u>x = 6²/₅            x = 0
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