Calculate $c\cdot\ln{\frac{4^{25}}{5^{16}}}$ Let $f(x)=\ln{x}$ and $g(x)=x^2$. Let $c$ be any constant given $c\in(4,5)$.

Calculate $c\cdot\ln{\frac{4^{25}}{5^{16}}}$


$c\cdot\ln{5}-8$
$c\cdot\ln{4}-8$
$2(c^2\cdot\ln{5}-8)$
$2(c^2\cdot\ln{4}-8)$



  Here is an attested
  graph showing that there are indeed solutions for $c\in(4,5)$.


My Attempt:

I know that I have to use Langrange's Mean Value Theorem.

So I started as,

$$c\cdot(5^2\ln{4}-4^2\ln5)$$

$$c\cdot5^2\cdot4^2(\frac{\ln{4}}{4^2}-\frac{\ln{5}}{5^2})$$

Then tried defining,

$$h(x)=\frac{f(x)}{g(x)}=\frac{\ln{x}}{x^2}$$

then by LMVT, there must be a $d\in(4,5)$ such that,

$$h'(d)=\frac{h(5)-h(4)}{5-4}$$

$$\frac{1-2\ln{d}}{d^3}=h(5)-h(4)$$

I can subsitute this in the real equation giving,

$$c\cdot4^2\cdot5^2(\frac{2\ln{d}-1}{d^3})$$

But the $5^2\cdot4^2$ part is giving me problem. I know I can eliminate this if I had dissolved them into the $h(x)$ in the original equation. But I was not able to form a $h(x)$ without taking $5^2\cdot4^2$ commmon.

What am I missing? 

Thanks for the help.
Because the question is unclear I'm going to try and articulate several versions of what I think the problem might be and then solve all of them.   

problem 1:
let $f(x)=\ln{x}$, and  $g(x)=x^2$ 

part 1:

construct a new function $h(x)$ by applying elementary functions$(+,-,\cdot,/)$ onto $f(x)$ and $g(x)$ (which can include the multiplying by a constant at the end) such that $h(5)-h(4)=\ln{\frac{4^{25}}{5^{16}}}$

part 2:

Using MLVT find the value for $c$ when $h’(c)=\frac{h(5)-h(4)}{5-4}$

solution: 

$$let: h(c)=\ln(4)\cdot \ln(5)\cdot \frac{g(c)}{f(c)}=\ln(4)\cdot \ln(5) \cdot \frac{c^2}{\ln(c)}$$

$$h’(c)=\ln(4)\cdot \ln(5)\cdot \frac{(2\cdot c\cdot \ln(c)-1)}{(\ln(c))^2}$$
$$\ln(\frac {4^{25}}{5^{16}})=\ln(4)\cdot \ln(5)\cdot\frac{(2\cdot c\cdot \ln(c)-1)}{(\ln(c))^2}$$
$$\ln(4^{25})-\ln(5^{16})=\ln(4)\cdot \ln(5)\cdot\frac{(2\cdot c\cdot\ln(c)-1)}{(\ln(c))^2}$$
$$25\cdot\ln(4)-16\cdot\ln(5)=\ln(4)\cdot \ln(5)\cdot\frac{(2\cdot c\cdot\ln(c)-c)}{(\ln(c))^2}$$
$$\frac{25}{\ln(5)}-\frac{16}{\ln(4)}=\frac{(2\cdot c\cdot\ln(c)-c)}{(\ln(c))^2}$$
$$(\frac{25}{\ln(5)}-\frac{16}{\ln(4)})\cdot(\ln(c))^2=2\cdot c\cdot\ln(c)-c$$
$$(\frac{25}{\ln(5)}-\frac{16}{\ln(4)})\cdot(\ln(c))^2-2\cdot c\cdot\ln(c)+c=0$$
$$c \approx 4.495$$
the reason why $h(x)=4^2\cdot5^2\cdot\frac{f(x)}{g(x)}$ doesn't work (which was essentially the OP's first attempt) is because the result of $h(5)-h(4)=4^2\ln{5}-5^2\ln4$ not $5^2\ln{4}-4^2\ln5$

problem 2 (this problem is assuming that d=c from the OP’s question. He said “I could state that d=c” in one of the comment threads of this web page)

$let: f(x) = \ln(x), g(x)=x^2$ Let $c$ be any constant given $c∈(4,5)$

part 1:construct a new function $h(x)$ by applying elementary functions$(+,-,\cdot,/)$ onto $f(x)$ and $g(x)$ (which can include the multiplying by a constant at the end) such that $h(5)-h(4)=\ln(\frac{4^{25}}{5^{16}})$

part 2: Given that $h’(c)=\frac{(h(5)-h(4))}{(5-4)}=\ln(\frac{4^{25}}{5^{16}})$ find when $c\cdot f’(c)=0$

the process and end result of this problem is almost identical to the last equation of problem 1. The last equation of the previous problem differs from the last equation of this problem on the last two terms on the left hand side of the last equation. These terms are multiplied by $c$.

$$(\frac{25}{\ln(5)}-\frac{16}{\ln(4)})\cdot(\ln(c))^2-2\cdot c^2\cdot\ln(c)+c^2=0$$
$$c \approx 2.118&lt;4$$
problem 3

find all values for $c$ in the equations below when $4&lt;c&lt;5$ 
    $$c\cdot\ln⁡5−8=c\cdot\ln⁡ (\frac{4^{25}}{5^{16}})$$
    $$c\cdot\ln⁡4−8=c\cdot\ln⁡ (\frac{4^{25}}{5^{16}})$$
    $$2(c^2\cdot\ln⁡5−8)=c\cdot\ln⁡(\frac{4^{25}}{5^{16}})$$
    $$2(c^2\cdot\ln⁡4−8)=c\cdot\ln⁡(\frac{4^{25}}{5^{16}})$$
In the graphs that the OP has posted only the third and the fourth equations intersect in the range $4&lt;c&lt;5$.
Equation 3
$$2\cdot(c^2\cdot\ln⁡5−8)=c⋅\ln⁡ (\frac{4^{25}}{5^{16}})$$
$$2\cdot c^2\cdot\ln⁡5−16=c\cdot\ln⁡ (\frac{4^{25}}{5^{16}})$$
$$2\cdot c^2\cdot\ln⁡5-c\cdot\ln⁡ (\frac{4^{25}}{5^{16}})−16=0$$
$$c≈4.0073$$
Equation 4
$$2\cdot(c^2\cdot\ln⁡4−8)=c\cdot\ln⁡(\frac{4^{25}}{5^{16}})$$
$$2\cdot c^2\cdot\ln⁡4−16=c\cdot\ln⁡(\frac{4^{25}}{5^{16}})$$
$$2\cdot c^2\cdot\ln⁡4-c\cdot\ln⁡(\frac{4^{25}}{5^{16}})−16=0$$
$$c≈4.4959$$
Based on the comments, you are asking if there is any $c\in (4,5)$ such that $c\cdot \ln \frac {4^{25}}{5^{16}}$ is equal to one of the following four expressions.  We can make a table of the values of the five at $c=4$ and $c=5$
$$\begin {array}{r|r|r} \text{Expression}&amp;c=4&amp;c=5\\ \hline 
c\cdot\ln \frac {4^{25}}{5^{16}}&amp;35.6254&amp;44.5318\\
c\cdot \ln (5)-8&amp;-1.5622&amp;0.0472\\
c\cdot \ln(4)-8&amp;-2.4548&amp;-1.0685\\
2(c^2\cdot\ln(5)-8)&amp;35.5020&amp;64.4719\\
2(c^2\cdot\ln(4)-8)&amp;28.3614&amp;53.3147
\end {array}$$
All of these functions are monotonically increasing.  We can see that the last two start below $c\cdot\ln \frac {4^{25}}{5^{16}}$ at $c=4$ and end above it at $c=5$.  The intermediate value theorem tells us that there is a point in the interval where they are equal.  If we want to plug in the constants we can find the values of $c$ explicitly as the equation is no worse than quadratic.  For example
$$c\cdot\ln \frac {4^{25}}{5^{16}}=2(c^2\cdot\ln(5)-8)\\
2\ln(5)c^2-c\cdot\ln \frac {4^{25}}{5^{16}}-16=0\\c\approx 4.0073$$
An exact form is available from Alpha, but I didn't find it enlightening.