log(2) + log(4) = log(2x)log(2 times 4) = log(2x)2 times 4 = 2 times 'x'x = 4
[log2 (x - 3)](log2 5) = 2log2 10 log2 (x - 3) = 2log2 10/log2 5 log2 (x - 3) = 2(log 10/log 2)/(log5/log 2) log2 (x - 3) = 2(log 10/log 5) log2 (x - 3) = 2(1/log 5) log2 (x - 3) = 2/log 5 x - 3 = 22/log x = 3 + 22/log 5
log x + 2 = log 9 log x - log 9 = -2 log (x/9) = -2 x/9 = 10^(-2) x/9 = 1/10^2 x/9 = 1/100 x= 9/100 x=.09
this is the question (log (base2) (x))^2- 12(log (base 2) (x)) + 32 = 0, I don't get this bit (log (base2) (x))^2, note the whole log is squared
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(x) + log(2) = log(2)Subtract log(2) from each side:log(x) = 0x = 100 = 1
log(2) + log(4) = log(2x)log(2 times 4) = log(2x)2 times 4 = 2 times 'x'x = 4
1
3^(-2x + 2) = 81? log(3^(-2x + 2)) = log(81) (-2x+2)log(3) = log(81) -2x = log(81)/log(3) - 2 x = (-1/2)(log(81)/log(3)) + 1
[log2 (x - 3)](log2 5) = 2log2 10 log2 (x - 3) = 2log2 10/log2 5 log2 (x - 3) = 2(log 10/log 2)/(log5/log 2) log2 (x - 3) = 2(log 10/log 5) log2 (x - 3) = 2(1/log 5) log2 (x - 3) = 2/log 5 x - 3 = 22/log x = 3 + 22/log 5
-2
log(36) = 1.5563To solve this problem without using a scientific calculator, factor 36 into 2*2*3*3, and use the formula:log(a*b) = log(a) + log(b)So, in this case:log(36) = log(2) + log(2) + log(3) + log(3) = 0.3010 + 0.3010 + 0.4772 + 0.4772 = 1.5564 (slight rounding error)
log x + 2 = log 9 log x - log 9 = -2 log (x/9) = -2 x/9 = 10^(-2) x/9 = 1/10^2 x/9 = 1/100 x= 9/100 x=.09
log base 2 of [x/(x - 23)]
this is the question (log (base2) (x))^2- 12(log (base 2) (x)) + 32 = 0, I don't get this bit (log (base2) (x))^2, note the whole log is squared
log x2 = 2 is the same as 2 log x = 2 (from the properties of logarithms), and this is true for x = 10, because log x2 = 2 2 log x = 2 log x = 1 log10 x = 1 x = 101 x = 10 (check)
-6