25^x=5^x+7

25^x=5^x+7

First we need to bring those x's to a more usable form.

Using logarithms, I can turn the x and x+7 into factors.

Log(A^B) = B * Log(A)

x * log(25) = (x + 7) * log(5)

Now re-arrange so all instances of x are on the same side of the equation.

^x/_(x+7) = ^log(5)/_log(25)

Now we got the annoying situation of x+7 being the denominator. I'm going to quickly flip the fractions around because we do not want an equation as the denominator.

^(x+7)/_x = ^log(25)/_log(5)

Let's quickly expand out our fraction to seperate the x's.

⁽⁷⁾/_x + ^(x)/_x = ^log(25)/_log(5)

Simplify...

1 + ⁷/_x = ^log(25)/_log(5)

Again, let's re-arrange to keep x seperate from everything else.

⁷/_x = ^log(25)/_log(5) + 1

Now, x is back on the denominator! Let's flip these fractions again to fix that (Note the "reciprocal" of 1 is 1, so that's why it won't change)

^x/₇ = ^log(5)/_log(25) + 1

Now we can remove the 7 by multiplying both sides by it.

x = (^log(5)/_log(25) + 1) * 7

And finally, let's work out these numbers.

x = ^{(log(5) * 7)}/_log(25) + 7 = 3.5 + 7 = 10.5

(Of course by the time I finish typing this, I'm a bit late at answering :P)

25^x=5^x+7

$$\begin{array}{rcl}
25^x &=& 5^x+7 \\
5^{2x} &=& 5^x+7 \\
5^{2x} - 5^x-7 &=& 0 \qquad |\qquad u = 5^x \\
u^2 -u-7 &=& 0\\
u_{1,2} &=& \frac{1}{2}\cdot [ 1\pm \sqrt{ 1-4\cdot (-7) } ] \\
u_{1,2} &=& \frac{1}{2}\cdot [ 1\pm \sqrt{ 29 } ] \\\\
u_1 &=& \frac{1+\sqrt{29}}{2} \qquad u_2 = \frac{1-\sqrt{29}}{2}\\\\
5^x&=& u \qquad |\qquad \ln() \\
x\ln{(5)} &=& \ln{(u)} \\
x &=& \frac{\ln{(u)}} {\ln{(5)}}\\\\
x_1 &=& \frac{ \ln{ ( \frac{1+\sqrt{29}}{2} ) } } {\ln{(5)}}\\
x_1 &=& 0.72126430677\\\\
x_2 &=& \frac{ \ln{ ( \frac{1-\sqrt{29}}{2} ) } } {\ln{(5)}} \qquad \text{no solution}\\
\end{array}$$

Very nice Heureka :)

If we accept Sir-Emo-etc's interpretation of the question we have:  5^2x = 5^x+7 so 2x = x+7 and x = 7.

You went awry after 1 + 7/x = log(25)/log(5) Sir-Emo..., you wrote 7/x = log(25)/log(5) + 1, but should have had 7/x = log(25)/log(5) - 1 

Also, you need to be careful when inverting both sides of an equation that you invert the whole of the side, not the individual terms!

.

Woops, thanks for correcting me. I'll try to avoid those mistakes in future.

And in honesty, my method was quite makeshift there. Kinda fiddled with the numbers way more than really was needed xD (And I completely forgot about how fractions worked. Gj me). I Much prefer Heureka's working (Particularly as "5^x+7" more likely means "(5^x)+7").

Hi anon,

I like people answering like you did, if you get involved it means you are much more likely to learn yourself, or maybe someone will learn from you. :)

I always think it is ultimately the askers decision to accept or reject an answer so if they copy your wrong answer they is not your fault.   

(This assumes everyone answers in good faith of course)