Question

The value of log°1.2bar to the base 3/4
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Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:

1,2618595071

Step-by-step explanation:

If you have enough time and like to solve artithmetical operations, you can begun with any fair approximation.

For example: 318=387,420,489≃4×108318=387,420,489≃4×108 .

And 321=10,460,353,203≃1010321=10,460,353,203≃1010 .

By taking base 10 logarithm on both sides:

318log31818log31818log34≃4×108≃log4×108≃8+log4≃8log3+log4log3≃8⋅2110+log34≃18−16810=1,2321log32121log3log3≃1010≃log1010≃10≃1021318≃4×108321≃1010log⁡318≃log⁡4×108log⁡321≃log⁡101018log⁡3≃8+log⁡421log⁡3≃1018≃8log⁡3+log⁡4log⁡3log⁡3≃102118≃8⋅2110+log3⁡4log3⁡4≃18−16810=1,2

Now. Remember that x=log34x=log3⁡4 means that 3x=43x=4 . We need to improve our precision.

Let’s take x=x0+dx0x=x0+dx0 , where x0x0 is a good approximation of xx , and dx0dx0 is the error. Then:

3x0+dx0=3x0⋅3dx03dx01+dx0ln3dx0ln3dx0=4=43x0≃43x0≃43x0−1=4−3x03x0≃4−3x03x0ln3if dx0≃0, then 3dx0≃1+dx0ln33x0+dx0=3x0⋅3dx0=43dx0=43x0if dx0≃0, then 3dx0≃1+dx0ln⁡31+dx0ln⁡3≃43x0dx0ln⁡3≃43x0−1=4−3x03x0dx0≃4−3x03x0ln⁡3

We still need ln3ln⁡3 , for which we take the approximation 33=27≃10e33=27≃10e , and therefore:

333log333−2110ln3=log3loge≃10e≃1+loge≃1log3+logelog3≃logelog3≃10933≃10e3log⁡3≃1+log⁡e3≃1log⁡3+log⁡elog⁡33−2110≃log⁡elog⁡3ln⁡3=log⁡3log⁡e≃109

So, in our formula, we get that:

dx0≃0.94−3x03x0dx0≃0.94−3x03x0

You can try several times.

If you begin with x0=1.2x0=1.2 and you keep iterating: dxn=0.94–3xn3xndxn=0.94–3xn3xn , and xn+1=xn+dxnxn+1=xn+dxn , then we can have the series

x0x1x2x3x4x5x6=1.2=1.2632898741=1.2618767079=1,2618597008=1,2618595093=1,2618595072=1,2618595071x0=1.2x1=1.2632898741x2=1.2618767079x3=1,2618597008x4=1,2618595093x5=1,2618595072x6=1,2618595071

This is a very good approximation