Question

Please if anyone can help me with this, i;m trying to learn how to do this type of a problem,

solve for x using logs
10^(x+8) = 6e^(3-x)

please show me how to solve this

Answer 1

take the natural log of both sides

10^(x+8) = 6e^(3-x)

ln[ 10^(x+8) ] = ln[ 6e^(3-x) ]

ln[ 10^(x+8) ] = ln(6) + ln( e^(3-x) )

(x+8)*ln(10) = ln(6) + (3-x)*ln( e )

(x+8)*ln(10) = ln(6) + (3-x)*1

(x+8)*ln(10) = ln(6) + 3-x

do you see how to finish up?

Answer 2

@jim_thompson5910, to be honest, not a clue how to finish it. keep coming up with the wrong answer

Answer 3

(x+8)*ln(10) = ln(6) + 3-x

x*ln(10)+8*ln(10) = ln(6) + 3-x

x*ln(10)+8*ln(10)+x = ln(6) + 3

x*ln(10)+x = ln(6) + 3 - 8*ln(10)

x(ln(10)+1) = ln(6) + 3 - 8*ln(10)

x = ( ln(6) + 3 - 8*ln(10) )/( ln(10)+1 )


So the exact answer is

\[\Large x = \frac{\ln(6) + 3 - 8\ln(10)}{\ln(10)+1}\]

Answer 4

awesome my brother figured it out

Answer 5

ok great

Answer 6

thank you so much!