Solve for $x$: $5^{x+2} + 2^{x+1} = 2^{x+5} + 13\cdot{5^x}$

Question

I just had a unit test on Logarithms and the thinking question was to Solve for x with the following equation given : $$5^{x+2} + 2^{x+1} = 2^{x+5} + 13\cdot5^x$$

The answer is $1$ but the question is worth 5 marks so how would you show your work for this question? I tried logging both sides but then I didn’t know got to expand after that. I think everyone got this question wrong in my class... so I hope she doesn’t count the question.

Answer 1

We have $$25\cdot5^x+2\cdot2^x=32\cdot2^x+13\cdot5^x,$$ collecting like terms $$12\cdot 5^x=30\cdot2^x,$$ dividing through by $6$

$$2\cdot 5^x=5\cdot 2^x,$$ or equivalently

$$\frac{2}{5}=\Big(\frac{2}{5}\Big)^x,$$

so of course

$$x=1.$$

Answer 2

write your equation in the form $$25+\left(\frac{2}{5}\right)^x\cdot 2=\left(\frac{2}{5}\right)^x\cdot 32+13$$ and Substitute $$t=\left(\frac{2}{5}\right)^x$$

Answer 3

$$5^{x+2}+2^{x+1}=2^{x+5}+13(5^x)$$$$5^25^x+2^12^x=2^52^x+13(5^x)$$$$(5^2-13)(5^x)=(2^5-2^1)(2^x)$$$$12(5^x)=30(2^x)$$$$2(5^x)=5(2^x)$$$$5^{x-1}=2^{x-1}$$$$\left(\frac52\right)^{x-1}=1$$$$(x-1)\ln\frac52=(x-1)\ln1=0$$$$x-1=0$$$$x=1$$

Answer 4

You actually don't need to use logarithms to solve the problem - just the laws of exponents. So we're given the equation:

$5^{x+2}+2^{x+1}=2^{x+5}+13\cdot5^x$

Now we rearrange the equation to get all the terms with bases of $2$ on one side, and $5$'s on the other side, which gives us:

$5^{x+2}-13\cdot5^x=2^{x+5}-2^{x+1}$

Remember the exponent law, $b^x\cdot b^y=b^{x+y}$, so we can rewrite the equation as,

$(5^x\cdot5^2)-13\cdot5^x=(2^x\cdot2^5)-(2^x\cdot2^1)$

Factoring, we get

$5^x(5^2-13)=2^x(2^5-2^1)$

$5^x(25-13)=2^x(32-2)$

$5^x(12)=2^x(30)$

$\frac{5^x}{2^x}=\frac{30}{12}$

$(\frac{5}{2})^x=\frac{5}{2}$

Now its clear that the solution to this problem is $x=1$. Hope that helps.