You do not. You divide by it.
For any two positive bases a and b, and any positive number x,
logax = logbx/logba.
Here b = e and a = 10
so
log10x = logex/loge10.
e2.303 = 10 (approx) so 2.303 = loge10 and so the result (though not as you stated it) follows.
To make a natural log a log with the base of 10, you take ten to the power of you natural log. Ex: ln15=log10ln15=log510.5640138 I'm sorry if you don't have a calculator that can do this, but this will work.
Due to the rubbish browser that we are compelled to use, it is not possible to use any super or subscripts so here goes, with things spelled out in detail: log to base 2a of 2b = log to base a of 2b/log to base a of 2a = [(log to base a of 2) + (log to base a of b)] / [(log to base a of 2) + (log to base a of a)] = [(log to base a of 2) + (log to base a of b)] / [(log to base a of 2) + 1]
18.057299999999998
By Euler's formula, e^ix = cosx + i*sinx Taking natural logarithms, ix = ln(cosx + i*sinx) When x = pi/2, i*pi/2 = ln(i) But ln(i) = log(i)/log(e) where log represents logarithms to base 10. That is, i*pi/2 = log(i)/log(e) And therefore log(i) = i*pi/2*log(e) = i*0.682188 or 0.682*i to three decimal places.
To enter a natural log, press the LN button. To enter a log with base 10, press the LOG button. To enter a log with a base other than those, divide the log of the number with the log of the base, so log6(8) would be log(8)/log(6) or ln(8)/ln(6). (The ln is preferred because in calculus it is easier to work with.)
To make a natural log a log with the base of 10, you take ten to the power of you natural log. Ex: ln15=log10ln15=log510.5640138 I'm sorry if you don't have a calculator that can do this, but this will work.
log0.1 50 = log10 50 / log10 0.1 ~= -1.699 To work out the log to any base b, logs to another base can be used: When logs are taken of a number to a power, then the power is multiplied by the log of the number, that is: log(bn) = n log b Taking logs to base b the power of b that equals the original number is being found, that is if: bn = m then logb m = n So, by using the logs to a base to which the answer can be known, the log to any base can be calculated: bn = m => n log b = log m => n = log m / log b => logb m = log m / log b as long as the same base is used for the logs on the right. It is normal to use base 10 or base e which are found on calculator buttons marked log (base 10) and ln (log natural - base e).
You will need 14 two's multiplied together to equal 16384. the answer to this can be found by log2(16384) = 14. Since most calculators don't have log base 2, you can do this: log(16384)/log(2) = 14. You can use the 'base 10' log or natural log [ln(16384)/ln(2) = 14]
Due to the rubbish browser that we are compelled to use, it is not possible to use any super or subscripts so here goes, with things spelled out in detail: log to base 2a of 2b = log to base a of 2b/log to base a of 2a = [(log to base a of 2) + (log to base a of b)] / [(log to base a of 2) + (log to base a of a)] = [(log to base a of 2) + (log to base a of b)] / [(log to base a of 2) + 1]
log 100 base e = log 100 base 10 / log e base 10 log 100 base 10 = 10g 10^2 base 10 = 2 log 10 base 10 = 2 log e base 10 = 0.434294 (calculator) log 100 base e = 2/0.434294 = 4.605175
log base 2 of [x/(x - 23)]
The log of infinity, to any base, is infinity.
log base e = ln.
It is the value that when the base you have chosen for your log is raised to that value gives 40,000 log with no base indicated means log to any base, thought calculators often use it to mean logs to base 10, which is often abbreviated to lg lg(40,000) = log{base 10} 40,000 ≈ 4.6021 ln(40,000) = log{base e} 40,000 ≈10.5966
18.057299999999998
The browser which is used for posting questions is almost totally useless for mathematical questions since it blocks most symbols.I am assuming that your question is about log base 3 of (x plus 1) plus log base 2 of (x-1).{log[(x + 1)^log2} + {log[(x - 1)^log3}/log(3^log2) where all the logs are to the same base - whichever you want. The denominator can also be written as log(3^log2)This can be simplified (?) to log{[(x + 1)^log2*(x - 1)^log3}/log(3^log2).As mentioned above, the expression can be to any base and so the expression becomesin base 2: log{[(x + 1)*(x - 1)^log3}/log(3) andin base 3: log{[(x + 1)^log2*(x - 1)}/log(2)
Log base 3 of 81 is equal to 4, because 3 ^ 4 = 81. Therefore, two times log base 3 of 81 is equal to 2 x 4 = 8.