Question about logarithmic eqations How to solve $4x+5^x=100$? I can't find how to solve it. I can't find a way to put the $x$'s into logarithmic form.
 A: As already said in comments and answers, this kind of equation cannot be solved in terms of elementary functions. As imulsion showed, there is a analytical solution in terms of Lambert function and for $ax+b^x=c$, the solution will be $$x=\frac{c}{a}-\frac{W\left(\frac{\log (b) b^{\frac{c}{a}}}{a}\right)}{\log (b)}$$
Otherwise, only numerical methods will be able to do it.
Let us consider $$f(x)=4x+5^x-100$$ By inspection, we easily find that the root is somewhere between $2$ and $3$. So, we can use something simple as Newton method which, starting from a "reasonable", will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ For this problem, let us start with $x_0=3$; then the method generates the following iterates $$x_1=2.819670304$$ $$x_2=2.788666164$$ $$x_3=2.787888138$$ $$x_4=2.787887664$$ which is the solution for ten significant figures.
But, if you plot the function, you will notice that it is quite stiff and more linear are the equation, faster we shall be able to solve it. So, we could rewrite the equation as $$g(x)=x\log(5)-\log(100-4x)$$ which will show the same zero as $f(x)$. The plot of $g(x)$ reveals almost a straight line.
Let us apply Newton method with $g(x)$ starting again with $x_0=3$; then the method generates the following iterates $$x_1=2.787915570$$ $$x_2=2.787887664$$ which is the solution for ten significant figures.
Quite faster, isn't it ?
You could notice that, being very lazy, starting with $x_0=0$, using Newton method for $g(x)=0$, the very first estimate will be $x_1=2.79196$ already quite close to solution. This does not imply that you have to be lazy !
A: It can't be solved using elementary algebra. You can plot it on a graph and find the point of intersection between $y = 5^x + 4x$ and $y = 100$, or you can use a numerical method to find an approximation. The Lambert W function can also be used, and Wolfram Alpha gives $$x = \frac{25 \log 5 - W(\frac{298023223876953125 \log 5}{4})}{\log 5} \approx 2.78789$$
