# Math Help - Transposition Question

1. ## Transposition Question

Hello first post.

I received a question as shown below but unfortuantely I keep on getting the answer as a Math Error on my calculator. I have tried the equation plenty of times but I keep on getting the same answer. Any help would be greatly appreciated!

p=-kvln(z0/v1)

Transpose for v1 and then evaluate when:

p=1.0008, k=1.23x10^-7, v=17.5 and z0 = 6x10^8

2. Originally Posted by RossF
Hello first post.

I received a question as shown below but unfortuantely I keep on getting the answer as a Math Error on my calculator. I have tried the equation plenty of times but I keep on getting the same answer. Any help would be greatly appreciated!

p=-kvln(z0/v1)

Transpose for v1 and then evaluate when:

p=1.0008, k=1.23x10^-7, v=17.5 and z0 = 6x10^8
Transforming the given equation:

$p=-k\cdot v \cdot \ln \left(\dfrac{z_0}{v_1}\right)$

$-\dfrac{p}{k\cdot v}=\ln \left(\dfrac{z_0}{v_1}\right)$ $~\implies~$ $e^{-\frac{p}{k\cdot v}} = \dfrac{z_0}{v_1}$ $~\implies~$ $\dfrac1{e^{\frac{p}{k\cdot v}}} = \dfrac{z_0}{v_1}$ $~\implies~$ $\boxed{v_1 = z_0 \cdot e^{\frac{p}{k\cdot v}}}$

Plug in the given values:

$v_1 = 6 \cdot 10^8 \cdot e^{\frac{1.0008}{1.23 \cdot 10^{-7}\cdot 17.5}}$

You'll get an exponent of 457508.5714... Which means that $e^{457508.5714}$ is much too large for your calculator. Therefore you get a MATH ERROR warning.

According to my computer the result is approximately 1.0325673*10^201933.

That is a number with nearly 202000 digits.

So I guess that there must be a small but very effective mistake in your calculations

3. Originally Posted by RossF
Hello first post.

I received a question as shown below but unfortuantely I keep on getting the answer as a Math Error on my calculator. I have tried the equation plenty of times but I keep on getting the same answer. Any help would be greatly appreciated!

p=-kvln(z0/v1)

Transpose for v1 and then evaluate when:

p=1.0008, k=1.23x10^-7, v=17.5 and z0 = 6x10^8
You will not get the answer if you punch the numbers straight into your calculator. This is to tell you education does make a whole world of a difference. Read your textbook before you attempt.

I suggest that you expand the RHS of the equation following the logarithm principle; then, get rid of the negative sign using the logarithm principle. When the first two steps are done, put your unknow on the LHS of the equation, and all knows on RHS. Last, punch the numbers into the calculation. Wallah, it's as easy as eating and drinking.

The ancient did not invent algebra using today's calculator. Use your Log table and throw away your calculator if it does you no good.

4. Thank you novice and earboth.

novice, I actually did carry out the calculations one-by-one rather than keying it all into the calculator. I made v1 the subject before using the calculator. I know (due to experience) mistakes can be made by just keying everything into the calculator without breaking it down. Unfortunately, we do not get log tables in the class, we are just expected to receive our answers of the calculator.

Anyway thanks for the help