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You complicated this equation. 1^x=1, so (8^x)²-(8^x)-1=0 Now solve for 8^x (which can't be negative) and then take logarithm on both sides to obtain the value of x.
Sorry my ignorance, but why log (root5 -1)/2 over log 1/8 is equal to log (root5 -1)/2 in base of 1/8? I mean, I was only familiar to the law that you could subtract these 2 logarithms Like log (root5 -1)/2 - log (1/8)
I think it's a bit easier to rewrite the equation as: (10⁰)ˣ + (10¹)ˣ = (10²)ˣ then (10ˣ)⁰ + (10ˣ)¹ = (10ˣ)² then (10ˣ)² - (10ˣ)¹ -1 = 0 Now, you have a quadratic in disguise with 10ˣ, which you solve using the quadratic formula. This gets you 10ˣ = (1+√5)/2 as you only take the positive solution. Log both sides and get your answer without the awkward base 1/10 log. Nice video though, your solution is very creative.
^= read as to the power *= read as square root As per question 2^a+4^a=8^a 2^a+(2^a)^2=(2^a)^3 Let 2^a=t So, t+t^2=t^3 t(1+t)=t^3 1+t=t^3/t 1+t=t^2 t^2-t-1=0 Here, a=1, b=-1, c=-1 D=*{(-1)^2-4.1.(-1)} =*{1+4)=*5 t=(-1±*5)/2 2^a =(-1±*5)/2 Take log log 2^a=log (-1±*5)/2 a. log2 ={log (-1±*5)}-log 2 a=[{log (-1±*5)}-log 2]/log2 (may be )
Why overcomplicate the first one so much? Just substitute for y=5^x and solve accordingly. You avoid negative logs and the whole first division part which doesn't really achieve anything.
Thanks for this. I love your clear reasoning and explanation. Wonder if an improvement would be to substitute u = 6^x? This avoids some of the fractions. Then, as 36^x = u^2x, you can write the quadratic as u^2 - u - 1 = 0. Solving this with quadratic formula gives u = { 1+ sqrt(5) } / 2 , plus a negative answer which can be rejected. 6^x = { 1+ sqrt(5) } / 2 log 6^x = log [ { 1+ sqrt(5) } / 2 ] x log 6 = log [ { 1+ sqrt(5) } / 2 ] x = log [ { 1+ sqrt(5) } / 2 ] / log 6 x = log6 [ { 1+ sqrt(5) } / 2 ] Also... { 1+ sqrt(5) } / 2 is the golden ratio... a nice little Easter egg they slipped in there.
Recognize this as a Golden Ratio problem for easy solution. Divide the given equation by 4^x and rearrange to (4^x)^2-4^x-1=0. Substitute Φ=4^x to obtain Φ^2-Φ-1=0, which has positive root Φ=(√5+1)/2 and solution x=lnΦ/ln4.
Dreadful presentation. Bad technique, bad English, and deceptive title. Anyone with basic math can see the real root instantly and mentally. This presenter is so sloppy he can't even be bothered to use fresh sheets of paper. YUCK! Beside the trivial real solution there are 8 complex solutions not mentioned. Among them are solutions that 8% are unlikely to get right, for example: x = -(3/2)^(2/3) sin(1/3 (tan^(-1)(sqrt(2)) - π)) - (3^(1/6) cos(1/3 (tan^(-1)(sqrt(2)) - π)))/2^(2/3) + i ((3/2)^(2/3) cos(1/3 (tan^(-1)(sqrt(2)) - π)) - (3^(1/6) sin(1/3 (tan^(-1)(sqrt(2)) - π)))/2^(2/3))