To solve this equation, we can first simplify it by using the property of logarithms that states (a^(log(a)) = a for a > 0:
(x+1)^(lg(x+1)) = 100(x+1)=> (x+1)^(lg(x+1)) = 100(x+1)^1=> (x+1)^(lg(x+1)) = (x+1)^2
Now we can equate the exponents since the bases are the same:
lg(x+1) = 2
Now, we can rewrite this logarithmic equation in exponential form:
10^(lg(x+1)) = 10^2
This simplifies to:
x + 1 = 100
Therefore, the solution to the equation (x+1)^(lg(x+1)) = 100(x+1) is x = 99.
To solve this equation, we can first simplify it by using the property of logarithms that states (a^(log(a)) = a for a > 0:
(x+1)^(lg(x+1)) = 100(x+1)
=> (x+1)^(lg(x+1)) = 100(x+1)^1
=> (x+1)^(lg(x+1)) = (x+1)^2
Now we can equate the exponents since the bases are the same:
lg(x+1) = 2
Now, we can rewrite this logarithmic equation in exponential form:
10^(lg(x+1)) = 10^2
This simplifies to:
x + 1 = 100
Therefore, the solution to the equation (x+1)^(lg(x+1)) = 100(x+1) is x = 99.