log3(x+1)=log9(26+2x)with calculation...

X?

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- Sep 14th 2009, 02:27 AMsf1903Simpel Log
**log3(x+1)=log9(26+2x)**with calculation...

X?

- Sep 14th 2009, 02:59 AMAmer
- Sep 14th 2009, 06:43 AMpacman
we may use this logarithmic identity by EULER, log (base b) a = (log a)/(log b)

solve for x: log3(x+1)=log9(26+2x)

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log (x + 1)/(log 3) = log (2x + 26)/(log 9) = log (2x + 26)/(2 log 3)

cross-multiply,

2 (log 3) log (x + 1) = (log 3) log (2x + 26), cancel log 3

2 log (x + 1) = log (2x + 26),

log (x + 1)^2 = log (2x + 26),

take anti-log og both sides,

(x + 1)^2 = 2x + 26,

x^2 + 2x + 1 = 2x + 26,

2x^2 - 2x + 2x - 26 +1 = 0,

x^2 - 25 = 0

factoring,

(x + 5)(x - 5) =0,

x = {-5, 5}, but only 5 is the valid answer

the graph below is wrong - Sep 14th 2009, 06:53 AMpacman
x^2 - 25 = 0, this is the graph

- Sep 14th 2009, 08:37 AMSoroban
Hello, sf1903!

Another approach . . .

Quote:

Solve for

Let: .

Square both sides: .

. . We have: .

The equation becomes: .

Exponentiate both sides: .

. . and so on . . .