PreCalc Help- Logarithms?

Question

Can anyone help me answer this problem?: When solving the equation 5^x=20, will you get the same answer when you use common logarithms as when you use natural logarithms? Why?... thanx in advance.

Answer 1

Yes, because to get the answer you first take the logarithm, solve for log(x) or ln(x), and then take the antilogarithm. Regardless of which base you use, you end up reversing it, so the answer will be the same either way.

Answer 2

Yes you will get the same answer no matter which logarithm you use. Because if you do the same thing to both sides of the equation both sides retain equal values.


Logarithms are the inverses of exponents, so a logarithm is always equal to the exponent. So by taking the log of a side you're raising it to a power. No matter what power/ logarithm you use, you'll always get an equivalent equation.

Answer 3

Another way to look at it is that a logarithm to any base can be converted to another base merely by multiplying by a constant. So if you solve the problem by taking logs of both sides you get

log(5^x) = log(20)

Whatever base those logs are, I can convert to another base by multiplying them by some constant k, so in another base the equation becomes

k*log(5^x) = k*log(20)

The constant k cancels out and does not change the result.

In particular, to change from natural to common logs the constant is log(e):

log(x) = log(e)*ln(x)

Answer 4

a million 2*ln x = ln(x^2), ln 4 + ln (2x + 3) = ln 4(2x + 3) 2 quickly replace e^x through y to grant you the quadratic equation y^2 - y - 12 ascertain that to get 2 y values then replace y through e^x. 3 2^x will cancel in the time of (allowed because it won't be able to be 0) to grant x^2 - 2x - 3 = 0 are you able to end each and each with those tricks?

Answer 5

Yes, you do get the same answers.

5^x = 20

log5 20 = x

x= log 20 / log 5


Same answers one is just a little more expanded.