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]
log37 - log3x = 4 log3(7/x) = 4 7/x = 34 = 81 x = 7/81
18.057299999999998
117 = 3 × 3 × 13245 = 5 × 7 × 7
log(5)125 = log(5) 5^(3) = 3log(5) 5 = 3 (1) = 3 Remember for any log base if the coefficient is the same as the base then the answer is '1' Hence log(10)10 = 1 log(a) a = 1 et.seq., You can convert the log base '5' , to log base '10' for ease of the calculator. Log(5)125 = log(10)125/log(10)5 Hence log(5)125 = log(10) 5^(3) / log(10)5 => log(5)125 = 3log(10)5 / log(10)5 Cancel down by 'log(10)5'. Hence log(5)125 = 3 NB one of the factors of 'log' is log(a) a^(n) The index number of 'n' can be moved to be a coefficient of the 'log'. Hence log(a) a^(n) = n*log(a)a Hope that helps!!!!!
Well first of all, you should know that anything raised to the 0th power is 1. So the answer is 0: 7^0=1. But if the answer's not so obvious, you should use logarithms. If 7^x=1, then the log (7) of x = 1. So you'd just find the inverse log (base 7) of 1. Some calculators can do that, some can't. The ones that can't usually can only find the log of a number in base 10 or base e (which is ~2.7182818), so you'd have to do a logarithmic conversion. To do a logarithmic conversion, you'd have to use the base conversion formula: log (base 7) of x = log (base 10) of x/log (base 10) of 7. Both sides of that equation equals 1 because of the information that you're giving me. So... log (base 10) of x/log (base 10) of 7 = 1. Using algebra, we're going to try to get x by itself on one side of the equation. We'll start by getting rid of the denominator by multiplying both sides by log (base 10) 7. The result: log (base 10) of x = 1 We'll isolate x from the logarithm by taking the inverse log of the left side. We'll take the inverse log of the right side to maintain the equality. x = log inverse (base 10) of 1. You can use a calculator for that. Well, you COULD, but you should have remembered that the log inverse of 1 with base ANYTHING is 0, right? x = 0. When you're trying to visualize what's going on, always remember that logarithms are exponents. If 10^2=100, then "2" in that previous equation is a logarithm; it's the base 10 logarithm of 100; log (base 10) of 100 = 2. It's basically a way of switching around the order that each number comes in. That may help you keep this new mathematical language in perspective.
39 expressed in base 10 vs. 117 expressed in base 8? 7 + 8 + 64 does not equal 39. Of course, you could get even more creative. Try base 16 for one or the other.
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
log37 - log3x = 4 log3(7/x) = 4 7/x = 34 = 81 x = 7/81
117= base* height 117/13=height 9 is d height
log base 2 of [x/(x - 23)]
The log of infinity, to any base, is infinity.
log base e = ln.
When the unknown is in the power you need to use logs (to any base) and the rule: log(a^b) = b × log(a) Thus: 10^x - 4 = 7 → 10^x = 11 → x log 10 = log 11 → x = log 11 ÷ log 10 If you use common logs (to base 10) then: lg 10 = 1 → x = lg 11 ≈ 1.04
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).
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