Posted by Ladybugtwo questions Reiny please help on .
1. solve the following logarithmic equation log_8(x+8)+log_8(x+7)=2
what is the exact solution
2.se the rational theorem to find all the real zeros of the polynomial function. use the zeros to factor f over the real numbers.
f(x)=4x^4+5x^3+9x^2+10x+2
a.find the real zeros of f. x=?
b.then use the real zeros to factor f. f(x)=?
please show work.

college algebra 
Reiny,
using rules of logs, your equation becomes
log_{8} ((x+8)/(x+7)) = 2
(x+8)(x+7) = 8^2 = 64
x^2 + 15x + 56  64 = 0
x^2 + 15x  8 = 0
x = (15 ± √257)/2
but x > 7 or else the log part would be undefined
so x = (15 + √257)/2 
college algebra 
Reiny,
2.
if rational roots exist, they must be ±1, or ±2 or ±1/2, ±1/4
we can rule out any positive x's since all the terms have positive and will never add up to zero
f(1) = 4  5 + 9  10 + 2 = 0 , yeahh
by synthetic division
4x^4+5x^3+9x^2+10x+2
= (x+1)(4x^3 + x^2 + 8x + 2) , again all positives in the 2nd factor so try x = 1, 2, 1/2, 1/4
x=1
f(1) = 4+1 8 + 2 ≠ 0
f(2) = 32 + ... ≠0
...
f(1/4) = ... 0
so (4x+1) is a factor
Using long algebraic division I ended up with
4x^4+5x^3+9x^2+10x+2 = (x+1)(4x+1)(x^2 + 2)
so there are 2 real roots or zeros:
x = 1 and x = 1/4