Question 161686
Find the nth term ({{{a[n]}}}):
{{{Log[2](8)}}},{{{Log[4](8)}}},{{{Log[8](8)}}},{{{Log[16](8)}}},{{{Log[32](8)}}},{{{Log[64](8)}}},{{{Log[128](8)}}},...
You can see that the base of the logarithms is always 2 to some power, and the power of 2 goes up like this, starting with the first term {{{a[1]}}}:{{{2^1 = 2}}},{{{2^2 = 4}}},{{{2^3 = 8}}},{{{2^4 = 16}}},{{{2^5 = 32}}},{{{2^6 = 64}}},{{{2^7 = 128}}},..., so for the nth term {{{a[n]}}} the base would be:{{{2^n}}}
Now you can write the nth term as:
{{{a[n] = Log[2^n](8)}}} Now if you rewrite this in exponential form, you'll get:
{{{(2^n)^a[n] = 8}}} Substitute {{{8 = 2^3}}} to get:
{{{2^((n*a[n])) = 2^((3))}}} Now since the bases (2) are equal, the exponents must be equal, so...
{{{n*a[n] = 3}}} Divide both sides by n to get:
{{{highlight(a[n] = 3/n)}}} This is the expression for the nth term expressed in {{{p/q}}} form.