Can someone please help me with Pre-Cal and show me the steps?

The problem is: 14^(a+8.5)+3=9. I'm not sure if I should subtract 3 on both sides and if I do what steps do I take to figure out what a is equal to?

Question

Can someone please help me with Pre-Cal and show me the steps?

The problem is: 14^(a+8.5)+3=9. I'm not sure if I should subtract 3 on both sides and if I do what steps do I take to figure out what a is equal to?

anonymous · Answer

The equation is $$14^{a+8.5}+3=9$$

amistre64 · Answer

subtract, then log ...

amistre64 · Answer

something, plus 3, is 9 ... we know that 6+3 = 9 so whatever a is, it has to make the term equal to 6 ...

amistre64 · Answer

14^(a+k) = 6

log both sides ...

log 14^(a+k) = log 6  now use log properties ....  log p^n = n log p

(a+k) log 14 = log 6

divide and subtract ...

anonymous · Answer

so even though the original equation didn't have logs I can still use them?

anonymous · Answer

@amistre64 thank you so much for providing the steps! I just want to make sure I got them correctly. 

Now I have a+8.5log14=log6... do i divide the logs?

amistre64 · Answer

yes, the logs just represent some numbers, log14 has a constant value, and so does log6 ... so they are just numbers at this point

anonymous · Answer

so i would just do 14/6 which is 2.33333 and then subtract 8.5 from that to figure out a?

amistre64 · Answer

logs are inverses of exponents ... just like subtraction is the inverse of addition.  when you have exponents, you can use logs to undo things

amistre64 · Answer

lets say log14 = s, and log6 = t

(a+8.5)log14 = log6

(a+8.5)s = t

a+8.5 = t/s

t/s is not 6/14,  it is log 6/ log 14,  or simply:$$log_{14}(6)$$

amistre64 · Answer

or another way to see it .. if this is to hard.  is to use the property:$$n^{log_n(k)}=k$$
$$14^{log_{14}(6)}=6$$therefore$$a+8.5=log_{14}(6)$$

anonymous · Answer

i see that makes more sense now that I see all of the little steps! so to figure out a from there i would solve logv14(6) and subtract 8.5 from whatever i get?

amistre64 · Answer

correct

anonymous · Answer

would my answer be correct if i got a= -7.821?

amistre64 · Answer

be careful that the solution is asking for an approximation, or an exact 

exact solution would keep it as   $a=log_{14}(6)-8.5$

approx solution would round $log_{14}(6)-8.5$ to some number of decimal places

anonymous · Answer

it says to solve and then round my answer to the nearest ten-thousandth... so how would ido that?

amistre64 · Answer

then approximation is fine http://www.wolframalpha.com/input/?i=log_%7B14%7D%286%29-8.5

amistre64 · Answer

the nearest ten thousandths is 4 decies

10 000
  ^^^^
   4 zeros, so 4th decimal place

anonymous · Answer

awesome! thank you so much for all of your help!! :)

amistre64 · Answer

youre welcome