1. ## algebraic transformations

Can anyone help me with these equations.

1.
make T the subject of the logarithmic relationship
M= V In T

2.
the following formula occurs in thermodynamics
Q= R0 In (V2/V1)
Transform to make V2 the subject.

3.
The ratio of belt tensions in a v-belt transmission is given by
R= γ0 / e^sinα
transform to make α the subject

2. Originally Posted by daniel123
Can anyone help me with these equations.

1.
make T the subject of the logarithmic relationship
M= V In T
"l"n, not "I"n!! I have seen that for year, and I have never understood why people write "In(x)" "l" for "l"ogarithm, "n" for "n"atural!

Any way, you solve an equation (make T the subject) by "un doing" what has been done to T. Here, if you were given T you would find M by doing two things: first take the logarithm of T, then multiply by V. To solve for T, do the opposite, in the opposite order. The opposite of "mutiply by V" is "divide by V" so you divide both sides by V: M/V= V ln(T)/V= ln(T). Now you do the opposite of "logarithm" which is the exponential: f(x)= ln(x) is defined to be the inverse function to $\displaystyle g(x)= e^x$. Take the exponential of both sides:
$\displaystyle e^{M/V}= e^{ln(T)}= T$. $\displaystyle T= e^{M/V}$.

2.
the following formula occurs in thermodynamics
Q= R0 In (V2/V1)
Transform to make V2 the subject.
Same basic idea: $\displaystyle Q= R_0 ln(V2/V1)$. Q/R= ln(V2/V1). $\displaystyle e^{Q/R}= V2/V1$, $\displaystyle V1e^{Q/R}= V2$.
3.
The ratio of belt tensions in a v-belt transmission is given by
R= γ0 / e^sinα
transform to make α the subject[/QUOTE]
$\displaystyle R= \frac{y_0}{e^{sin(\alpa)}}$
Multiply both sides by $\displaystyle e^{sin(\alpha)}$:
$\displaystyle Re^{sin(\alpha)}= y_0$
Divide both sides by R:
$\displaystyle e^{sin(\alpha)}= \frac{y_0}{R}$
Take the logarithm (inverse to exponential) of both sides:
$\displaystyle ln(e^{sin(\alpha)}= sin(\alpha)= ln(y_0/R)$
Take the inverse sine of both sides:
$\displaystyle sin^{-1}(sin(\alpha))= \alpha= sin^{-1}(ln(y_0/R))$