I tried but I stuck when I simplified all the terms to 5's power of log base5 two and something~~ could someone help me ?5^2（log5 (2+X)）-2=5^(X+log5(2)) solve for x.

Question

dumbcow · Answer

$$\large 5^{2 \log_{5}(x+2)} -2 = 5^{x+\log_{5}2}  $$
?

anonymous · Answer

Ya that's the question, sorry that I've no idea how to type those correctly~~

dumbcow · Answer

ok lets simplify using some exponent properties
$$a^{mn} = (a^{m})^{n}$$
$$a^{m+n} = a^m *a^n$$
$$\rightarrow \large (5^{\log_{5}(x+2)})^{2} -2 = 5^{x}* 5^{\log_{5}2} $$

dumbcow · Answer

$$b^{\log_{b} x} = x $$
when base of log matches exponential base, they cancel each other out

dumbcow · Answer

$$\rightarrow (x+2)^{2} -2 = 2*5^x$$

anonymous · Answer

Oh I get it. Thank you sooooo much!

dumbcow · Answer

your welcome...but just so you know, this  cant be solved algebraically because "x" is both an exponent and part of polynomial

anonymous · Answer

mmhmm, I gonna try to work it out~~ I'm was sure but now I'm not very sure.

dumbcow · Answer

what class are you taking? 
you cant work it out because there is no way to isolate "x"
by looking at it and checking....i see that x=0 is a solution

anonymous · Answer

I'm taking IB HL maths. It's possible to solve because the exponent suppose to be 2, it's 5(log5(2+x)) instead of the other way, sorry that was my bad, I didn't have a good look at it. Now it's so much easier. Thank you very much!!!!!!