You take the log of both sides: log(2^X)=log(128) Then you use the property that log(a^b)=b log(a) The remaining steps should be apparent.
2^{x} = 128 the first thing you have to do is get both of the bases the same so in 2^{x}, x is the exponent (or power if you like) and 2 is the base. So what you have to do is get 128 with a base of 2. So, 128 = 2 * 2 * 2 * 2 * 2 * 2 * 2 128 = 2^{7} There for, 2^{x} = 2^{7} So, x = 7 This way may seem long, but take the example of 2^{2(x+1)} = 1024 1024 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 2^{10} 2^{2(x+1)} = 2^{10} 2(x+1) = 10 2x + 2 = 10 2x = 8 x = 4
there are also other cases where you have to take logs of both parts, like it initially was suggested. Say you have: [tex]2^{x+1}=35[/tex] you defenitely cannot express 35 as a power of 2, so in these cases you need to take the log of both parts, and choose a base that will be easier to work with. [tex]log_a 2^{x+1}=log35=>(x+1)log_a 2=log_a35=>x+1=\frac{log_a35}{log_a2}[/tex]
Yes it is - 1, you can do it in your head in the time it takes to find the log button on the calculator or if you happen to be an android 2, multiplying by 2 is a single shift instruction in a register which you can do in a single clock cycle, it would take at least 7 clock cycles to load the instruction into the FPU and get the answer back - especially if you have a long pipeline.
Too Simple! find the prime factorization of 128. Now, what is the exponent? 128 = 2 * 64 = 2 * 2*2*2 * 2*2*2 Count the factors of 2 in the factorization of 128.
mgb_phys way is definitely quicker. Note when we take logs of both sides, all we achieve is [itex]x= log_2 (128)[/itex], and to actually evaluate that we must work out [itex]2^7 = 128[/itex] anyway.