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Log31 y⁹ √odd graph b1

Why would any regular person need to know how to do this? How could it be used and applied to real life??

Log b function show enigma +- square roots function s quotation 36 hours later - minus root function y valid y++ Safe mode function c))

M/nnn √rt3 blog a .. q. Egg note q 1/1ac. U7-y

Ans : -(1/3)×[{log(*5-1)- log2}/(log2)] =-(1/3)[{log(*5-1)/log2}-1]...May be Explain later......

I would have gone straight to base 7 on the valid solution for logs. Reduces unnecessary steps and automatically cancels the exponential.

{12+12}=24 4^6:4^2^3:2^2^1^3 1^2^1^3 23(x ➖ 3x+2).

X is 2?

Answer is k=3/2.

Just simply m²+7m+50 = 0 m²+7m+12,25 = -37,75 (m+3,5)² = -37,75 m= -3,5±i√37,75 We can simplify furthermore but I don't see any sense

k=4 log2b3

{9+3}=12 3^9 3^3^2 1^3^1 3^1 (x ➖ 3x+1).

{8.4+8.4}=16.8 8^8.8 2^3^2^3.2^3 1^1^1^1^1^1.1^1^2 1^2 (k ➖ 2k+1).

12 😊😊🎉

what if i got k = log14 base 3 is it still correct?

{15x+15x ➖ }+{30x+30x ➖ }={30x^2+60x^2}=90x^4 30^60x^4;3^2^3x^2^2 1^1^3x^1^2 3x^2(x ➖ 3x+2).

8:41 unfortunately your handwriting is horrible, messy, untidy.

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love the video. keep up the good work :D

1/6

you can also continue using 5*i^2 instead of -5 and dont reject the solution

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Ofcourse some thing to learn.

x=sqrt6+sqrt7

Where’d you get the rule that (a^a = n^n) => (a = n) when either is known to be >= 1?