Alan

Here's a graph (drawn to scale) to illustrate Chris's answer:

This was given as a formula to find a distance. m-M=5log(d/10)

I went ahead and plugged in the given information, which led to it being 1.08=log(d/10) and I am still trying to find d, the distance. Any better?

Yes, much better!  there is no ^ here. So using a property of logarithms (see the Formulary) we can write:

1.08 = log(d) - log(10)

Now log to the base 10 of 10 is just 1 (ie log₁₀(10)=1) so 

1.08 = log(d) - 1 

Add 1 to both sides

2.08 = log(d)

Now raise both sides to the power 10

10^2.08 = d because 10^log(d) = d

So:

$${\mathtt{d}} = {{\mathtt{10}}}^{{\mathtt{2.08}}} \Rightarrow {\mathtt{d}} = {\mathtt{120.226\: \!443\: \!461\: \!741\: \!290\: \!6}}$$

 or d is approximately 120.

This still isn't meaningful!

What does the expression log^(d/10) mean? You have to have the logarithm of something before you can start raising to a power.  

Perhaps you mean log(d^1/10) in which case your equation would be 1.08 = log(d^1/10), which is equivalent to 1.08 = (1/10)log(d), so log(d) = 10.8 so d = 10^10.8. 

or do you mean 1.08 = log(d/10) in which case 1.08 = log(d) - log(10) so 1.08 = log(d) - 1 so log(d) = 2.08 so d = 10^2.08. 

or do you mean 1.08 = (log(d))^1/10, in which case 1.08¹⁰ = log(d) so d = 10^{(1.08^10)}?

You could have log(m^d/10) or log(m)^d/10, but you would then need to specify the value of m in order to find d.

Just type this expression, exactly as you have written it, directly into the calculator on this site (go to Home to find it).

Yes, but you would have to type in the relevant equations yourself.  The calculator here doesn't have them built-in.

We can do this as follows: (Note: In copying the LaTeX from Tekmaker and pasting it here the third equation has been renumbered to (1)!!!. The third equation below should be numbered (3), of course.)

$$Let's first rewrite the two given equations as: \begin{equation} y1 = -\frac{3}{2}x+\frac{5}{2} \end{equation} \begin{equation} y2 = -4x+5 \end{equation}$$

$$and the line perpendicular to the first of the above as: \begin{equation} y3 = mx+c \end{equation} where $m$ is its gradient and $c$ is its intersection with the y-axis.\\\\$$

$$The gradient, $m$, is given by -1/the gradient of the line to which it is perpendicular, so $m=\frac{2}{3}$\\\\ To find the point of intersection of lines (1) and (2) we need to equate them:$$ -\frac{3}{2}x+\frac{5}{2}=-4x+5 $$and solve for $x$. This is straightforward and results in $x=1$. Plugging this back in either equation (1) or equation (2) we find that the value of y at the intersection point is also $y=1$\\\\ Now we put the known values of $x$ and $y$ at the intersection point, together with our value for $m$ into equation (3):$$ 1=\frac{2}{3}*1+c$$

$$from which we get $c=\frac{1}{3}$; so, using equation (3), we get the equation of the perpendicular line as:$$ y3=\frac{2}{3}x+\frac{1}{3}

Let's plot a graph to see if the result looks sensible:

Actually kitty, the general convention is that √9 is just 3 not -3.  If we have an equation like x² = 9 then there are two possible values of x, namely ±√9.  However, the √9 part is just 3 {otherwise we'd have the situation that x = ±(±3), or even worse the infinite regression: x = ±(±(±(±......(±√9)....))) !!!!!}

Area of cross-section times length.  For a more specific answer you'd need to specify the shape of the wedge (with some values or parameters describing its various dimensions). 

I'm afraid neither your question nor your answer makes any mathematical sense.  What do you mean by log^d? Do you just mean log(d)?  Do you mean log(d/10)? Do you mean log(d)/10? Do you mean something else?  

And just because your log might be to base 10, that doesn't mean you can just cancel it with another 10.  

No, but 2π radians make a complete circle (i.e. is the equivalent of 360°).  One radian is the angle subtended at the centre of the circle by an arc of the circle such that the arc has the same length as the radius of the circle.  One radian is 360/2π degrees:

$${\mathtt{radian}} = {\frac{{\mathtt{360}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}} \Rightarrow {\mathtt{radian}} = {\mathtt{57.295\: \!779\: \!513\: \!082\: \!320\: \!9}}$$  degrees

 Note: My "No" refers to the question, not Melody's answer! Melody must have posted while I was writing.