SOLUTION: Solve for x log(base5)x+log(base5)(x+1)=7 I tried solving this equation but couldn't find the answer. Here are my step. log(base5)x+log(base5)(x+1)=7 log(base5)x(x+1)=7

Question 998769: Solve for x
log(base5)x+log(base5)(x+1)=7
I tried solving this equation but couldn't find the answer. Here are my step.
log(base5)x+log(base5)(x+1)=7
log(base5)x(x+1)=7
log(base5) x^2+x=7
5^7=x^2+x --- I am stuck here

Found 2 solutions by josgarithmetic, Boreal:
Answer by josgarithmetic(39618) (Show Source): You can put this solution on YOUR website!
Your steps mostly show non-standard notation; otherwise, you stopped too early without doing what you already know how to do.

Do like this:
As plain text, log(5,(x))+log(5,(x+1))=7
As rendered to look like standard notation,

Property of logs,
Distribute inside the log input,

So far what you would have shown better if done on paper. Why did you just stop there?

-----Quadratic Equation
------solutions in raw form, even though not very pretty with the shown as part of the expression.

-----the square root part becomes very near to 559, but...
THE prime factorization of 312501 is 3*7*23*647, so ...
YES, use .

x=279 or x=-280
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
keep going
5^7=78125
x^2+x-78125=0
quadratic formula, positive root only
x=(1/2)(-1+/- sqrt (1+78125*4); the radical is 559.018
divided by 2, x=279.0089 (do not forget the minus 1 for minus b. You need it)
log 5(279.0089)=3.4989
log 5 (280.0089)=3.5011
Those aren't exact, but they are extremely close, and they add to 7.
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