good job ! Another way was how you started at first 4*(4**x) = 512, then /4 gives 4**x = 512/4, then take logs of both sides gives x log 4 = log128, then x = log 128/log 4 ; gives x = 3.5. InventPeaceNotWar
Im not really good at math, and solved it basically like this in my head: 4 * 4^x = 512 so 4^x gotta equal 128, now 4^3 is 64 and 4^4 is 256 thus it has to be 3.5 to get the 128 thus x is 3.5 Is that basically what logs does too?
Well, log4(128) is 3.5 and it’s not so simple since it’s a fractional power and you really need to know that since 2 is a square root of 4, it’s the same as 4^0.5, that means that adding 0.5 to the 4’s power exponent is the same as multiplication by 2.
Your second question, log 10 instead of reducing to 1, you should split it into log 2x5. since yourr denominator is log 5, it is equal to a change of base. so log 10 = log base 5 5 + log base 5 2, so the final answer should be 3 + log base 5 2.
I'm a pragmatist. I saw the reduction to 4^x=128 which, while that can be solved with log...but why? Simple exponentiation gets us to x=4 quicker than i can grab my calculator. I didn't watch the video...not sure if that's what they did.
good job ! Another way was how you started at first 4*(4**x) = 512, then /4 gives 4**x = 512/4, then take logs of both sides gives x log 4 = log128, then x = log 128/log 4 ; gives x = 3.5. InventPeaceNotWar
Im not really good at math, and solved it basically like this in my head:
4 * 4^x = 512 so 4^x gotta equal 128, now 4^3 is 64 and 4^4 is 256 thus it has to be 3.5 to get the 128 thus x is 3.5
Is that basically what logs does too?
trial and error. x=3 is too small, x=4 is too big, so I tried x=3,5 and it worked. Maybe it was luck, but it only took ~10 seconds.
4*4^x=512 (2^9) -> 4^x=512 (2^9) /4(2^2) -> 4(2^2)^x=2^(9-2)=2^7 -> 2^2x=2^7 -> 2x=7 -> x=3,5
Well, log4(128) is 3.5 and it’s not so simple since it’s a fractional power and you really need to know that since 2 is a square root of 4, it’s the same as 4^0.5, that means that adding 0.5 to the 4’s power exponent is the same as multiplication by 2.
Your second question, log 10 instead of reducing to 1, you should split it into log 2x5. since yourr denominator is log 5, it is equal to a change of base. so log 10 = log base 5 5 + log base 5 2, so the final answer should be 3 + log base 5 2.
Third question, need not use quadratic equation, simple factoring will do.
I'm a pragmatist. I saw the reduction to 4^x=128 which, while that can be solved with log...but why? Simple exponentiation gets us to x=4 quicker than i can grab my calculator. I didn't watch the video...not sure if that's what they did.
But it’s not 4.
4^x = 128
xlog4 = log128
X = log4 + log32 / log4
X = 1+ log4 + log8 / log4
X = 2+ log4 + log2 / log4
X = 3+log2/log4
X = 3,5
=> 4×4ⁿ=512
=> 4ⁿ=512/4
=> 4ⁿ=128
=> 4ⁿ=2⁷
=> (2²)ⁿ=2⁷
=> 2²ⁿ=2⁷
=> 2n=7
=> n=7/2
=> n=3,5