If a^b equals (a^x)/10, will b equal x/10?

Thanks, Bertie.....I see where I made my error  (being in too much of a hurry!!)

Let's take it from here.....

log (10)(a^b) = log a^x       by a property of logs, we have

log 10 + log a^b = log a^x   and we have, using another log property

1 + (b) log a = (x) log (a)    

(b) log (a) = [ (x) log (a) - 1]     divide both sides by log (a)

b = [x - (1 / log (a)) ]    ≠   x/10

Thanks again, Bertie, for your sharp eyes!!!....I forgot that the "b" exponent wasn't being applied to the "10" (I hope I got it right this time !!!)    

If a^b equals (a^x)/10, will b equal x/10?

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Let's suppose that it's true    multiply both sides by 10    we have

10a^b = a^x    take the log of both sides

log 10a^b = log a^x       by a property of logs, we have

b log 10a = x log a      and using another log property, we have

b [log 10 + log a] = x log a        and because log10 = 1    we have

b [1 + log a] = x log a        solving for b, we have

b = [x log a] / [1 + log a]   which is definitely    ≠   x/10

Sorry CPhill, but you need to look again at that fourth step.