To solve ( \log_2 8 + \log_2 32 ), you can use the property of logarithms that states ( \log_b a + \log_b c = \log_b (a \cdot c) ). First, calculate ( \log_2 8 ), which is 3 because ( 2^3 = 8 ), and ( \log_2 32 ), which is 5 because ( 2^5 = 32 ). Therefore, ( \log_2 8 + \log_2 32 = 3 + 5 = 8 ). Alternatively, you could find ( \log_2 (8 \cdot 32) = \log_2 256 = 8 ) directly.
How do you solve 4y plus x equals 8
Your calculator won't usually have a function to calculate logs in base 5 or base 8 directly, but this can easily be solved. For example: log5125 = log 125 / log 5 (taking both logs in base 10, or both logs in base e) In this particular case, you can also solve the equation mentally - you don't even need a calculator! Just use the definition of a log: "To what power must I raise 5 to get 125?" The answer to this is, by definition, log5125. Similarly with log28.
It is: 8-2x
8x2 + 16x + 8 = 0
(7)-(-3)+(16)-(8) = 18
log2(8) = 3 means (2)3 = 8
How do you solve 4y plus x equals 8
Your calculator won't usually have a function to calculate logs in base 5 or base 8 directly, but this can easily be solved. For example: log5125 = log 125 / log 5 (taking both logs in base 10, or both logs in base e) In this particular case, you can also solve the equation mentally - you don't even need a calculator! Just use the definition of a log: "To what power must I raise 5 to get 125?" The answer to this is, by definition, log5125. Similarly with log28.
7Because it is a logarithmic sequence:2^0 =1 so log2 1 = 02^3 = 8 so log2 8 = 32^5 = 32 so log2 32 = 52^7 = 128 so log2 128 = 7
It is: 8-2x
-2
log3 81 × log2 8 × log4 2 = log3 (33) × log2 (23) × log4 (40.5) = 3 × (log3 3) × 3 × (log2 2) × 0.5 × (log4 4) = 3 × 1 × 3 × 1 × 0.5 × 1 = 9 × 0.5 = 4.5
8
8x2 + 16x + 8 = 0
(7)-(-3)+(16)-(8) = 18
It is: 12-8+5 = 9
5