How do I calculate with logarithms?

What you need:

• Basic knowledge of "potencies"

Logarithm - what is it?

• The logarithm is the inverse of the exponential function. But this definition usually doesn't help. It is easier to understand logarithms if you know that when calculating with them, the question of the exponent always arises.
• For example, if you have equation 2^x = 16, the solution for x can easily be found by raising the power accordingly, namely x = 4. In the logarithmic notation, x = log now applies₂ 16 (read: log of 16 to base 2). When you ask about the logarithm, you are always looking for the exponent with which you have to raise the base (here 2) to the power to get the value (here 16).
• If this is still a bit too theoretical for you, think back to the roots. This arithmetic operation was also the reverse of squaring (or exponentiation, if you have higher root regard). The logarithm is to be understood in this sense.

Calculating with logarithms - some examples

Against this mathematical background, some examples should explain how you can calculate with logarithms. In general, it helps (if you are not allowed to use a TR) to rewrite the logarithmic problem as an exponential problem:

• Wanted is log₃ 81, the exponent of x, with which you have to raise the base 3 to the power to get 81. Simply trying it out shows that x = 4, because 3⁴ = 81.


Pocket calculator: Enter the logarithm on the TI 30 eco RS

A pocket calculator, including the TI 30 eco RS, is only ever as smart as the one who ...

• A little more complicated to find is log_⁸ 2. In this case, 8 applies^x = 2, an initially somewhat astonishing task. Remember the calculations of the roots and you will find x = 1/3 because 3. Root (8) = 2.
• Also the Logarithmic Laws can help to solve problems in which a logarithm occurs. For example, do you need to log the task_a (a⁵), then you "pull" the power in front of the logarithm. The following applies: log_a (a⁵) = 5 * log_a (a) = 5 because a¹ = a.
• A special "delicacy" is of course Equationsin which the basis is sought, such as log_x 27 = 3. Here, too, you shouldn't get confused and set up the associated inverse equation. It is x³ = 27 (note that the unknown x is the base here) and you can easily find x = 3.