How do you solve #(log x)^2+ 3 log x - 10 = 0#?

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Answer 1 · 2016-03-02T15:29:52

#x=100# and #x=10^-5#

From the given:

#(log x)^2+3*log x -10=0#

this is a quadratic equation in log x

Let y=log x

then

#y^2+3y-10=0#
solve by factoring

#(y-2)(y+5)=0#

we have #y=2# and #y=-5#

Therefore

#log x=2#

#10^2=x#
#x=100#
~~~~~~~~~~~~~~
Also #y=-5#

and #log x =-5#

#10^-5=x#

#x=1/100000=0.00001#

#x=0.00001#

God bless...I hope the explanation is useful