• January 7th 2013, 08:47 AM
Please say me where I am wrong.... or show me how to continue to solve this equation...
Thanks
Attachment 26502
• January 7th 2013, 09:05 AM
jakncoke
is the comma supposed to be a dot in $2 lg^3 x - 1,5x$
• January 7th 2013, 09:12 AM
isn't it the same? you can write it like that: $\left( {2{{\lg }^3}x - {3 \over 2}x} \right) \cdot \lg x = {1 \over 2}$
• January 7th 2013, 03:56 PM
skeeter
if I remember correctly, the log symbology lg indicates a base 2 logarithm ... ?

... your work shows you are assuming it's a base 10 logarithm.
• January 8th 2013, 02:19 AM
in the country I live in, lg means log base 10, but I solved it already so no problem any more, it was a mistake in the textbook, there should be 1.5logx not just 1.5x
• January 8th 2013, 06:19 AM
HallsofIvy
Getting back to the original equation, because you have the unknown, x, as base and in the exponent, there is not going to be any "algebraic" method. You might, with a lot of algebraic manipulation, get this into a form in which you could apply "Lambert's W function" (defined as the inverse function to $f(x)= xe^x$) but I don't see any simple way to do that.