How do you solve log equations?

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SUMMARY

This discussion focuses on solving logarithmic equations, specifically the equations 3log(x-15) = (1/4)x and another unspecified equation. The participants highlight the challenges of isolating x and suggest that the logarithm is likely base e (natural logarithm). They conclude that the first equation has a solution at x=3, while the second equation requires numerical methods for solutions, particularly noting that log(x-15) is only valid for x > 16.

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Homework Statement



Solve:
a)
54_a334e0e5e62f9b16841565a8ba38fd98.png


b) 3log (x-15) = (1/4)x

The Attempt at a Solution


For a)
[URL]http://i1141.photobucket.com/albums/n596/physics-/43_6051b53e5bf48508fa9854606102060a.png[/URL]
[URL]http://i1141.photobucket.com/albums/n596/physics-/46_ab0ccd938c1710d9cb5bbcd6193d4eac.png[/URL]
[URL]http://i1141.photobucket.com/albums/n596/physics-/32_b606b77ac87094c3cb6471d0dffad16e.png[/URL]
And then I got stuck for how to isolate x.

For b)
I'm not sure how to deal with a variable exponent.
 
Last edited by a moderator:
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i hate to say it, but you might just have to multiply out the (4-x)^7. the expansion is [PLAIN]http://i1204.photobucket.com/albums/bb414/aeekey/equation.gif, which i got using http://wolframalpha.com" . the site is useful for straight computations or pesky expansions, but i wouldn't rely on it for everything.

for part b, I'm assuming the log is base e (ln) and not base 10. see what you can do with e^x = 12log(x-15), hm?
 
Last edited by a moderator:
Were you expected to solve these numerically? Because you can't solve them algebraically.

For
a) there is a nice value of x=3 that works, but the other real roots are going to be nasty irrationals that are the roots of that polynomial ptolema has shown.

b) 1/4x is positive for all x, and approaches zero for positive values of x, log(x-15) is only positive for x>16 so the solution is going to be 16 + a value approximately equal to 1/416 which is obviously tiny. You can only get a numerical solution for this equation as well.
 

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