<PRE><B>Question 24059</B>
&lt;p&gt;Sorry, about the slightly messy notation, but I&#39;m still trying to figure out how to use this thing.&lt;/p&gt;

&lt;p&gt;Ok, I assume the thing that&#39;s making it tricky for you is the rather nastly looking {{{lg[x]}}}. When I first saw this problem I thought eugh! What on earth am I going to do with that. I then remembered somebody&#39;s (probably not) famous theorem.&lt;/p&gt;

&lt;p&gt;Logs of different bases differ by a multiplicitive constant!&lt;/p&gt;

&lt;p&gt;Let me demonstrate this with an example, say we wanted to solve {{{14^x=29.6}}}. I&#39;m sure you could tell me straight away that the answer is {{{x=lg[14](29.6)}}}. But let&#39;s try and work out the answer another way. Lets take {{{lg[10]}}} of both sides of the equation.&lt;/p&gt;

{{{lg[10](14^x)=lg[10](29.6)}}}

&lt;p&gt;Applying one of the many &quot;laws of logs&quot;&lt;/p&gt;

{{{x*lg[10](14)=lg[10](29.6)}}}

&lt;p&gt;And finally dividing&lt;p&gt;

{{{x=lg[10](29.6)/lg[10](14)}}}

&lt;p&gt;So, we&#39;ve solved for {{{x}}} in two different ways, and thus we have worked out that&lt;/p&gt;

{{{lg[14](29.6)=lg[10](29.6)/lg[10](14)}}}

&lt;p&gt;Now if you can do the same thing and turn the nasty {{{lg[x]}}} into a much nicer {{{lg[3]}}} you should be able to do the problem.&lt;/p&gt;

&lt;p&gt;There is another possibly tricky bit, but see how you get on. please write back if you didn&#39;t follow that, or you get stuck further on.&lt;/p&gt;

&lt;p&gt;Hope that helps&lt;/p&gt;

&lt;p&gt;Kev&lt;/p&gt;</PRE>