1. ## Basic exponents

3^x + 5^x = 12

I have taken logs on both sides, used the power rule to get

xln3 + x ln5 = ln12

factored out x on the LHS to give x(ln3 + ln5) = ln12

combined logs

x(ln15) = ln12

divided through by ln15

x = (ln12)/(ln15)

The answer I get doesnt match what I want. Where have I gone wrong please?

2. Originally Posted by 200001
3^x $\color{red}+$ 5^x = 12
I fear that this equation cannot be solved symbolically.

I have taken logs on both sides, used the power rule to get

xln3 + x ln5 = ln12
That would be correct, if the original equation had been $3^x\cdot 5^x=12$.

factored out x on the LHS to give x(ln3 + ln5) = ln12

combined logs

x(ln15) = ln12

divided through by ln15

x = (ln12)/(ln15)
That's correct, provided, as I wrote above, the original equation really was $3^x\cdot 5^x=12$.

The answer I get doesnt match what I want. Where have I gone wrong please?
Either you have gone wrong in the very first step when you assumed that $\ln(3^x+5^x)=x\ln 3+x\ln 5$, or you have gone wrong in believing that the solution that you have found is not correct (or, rather, that it is "not what you want").

I run it through wolfram and their answer was the same as the given answer.

&#x28;3&#x5e;x&#x29;&#x2b;&#x28;5&#x5e;x&#x29;&#x3 d;12 - Wolfram|Alpha

How would I approach this?

4. Originally Posted by 200001
I run it through wolfram and their answer was the same as the given answer.

&#x28;3&#x5e;x&#x29;&#x2b;&#x28;5&#x5e;x&#x29;&#x3 d;12 - Wolfram|Alpha

How would I approach this?
Use the solution in Wolframalpha as a hint: "Numerical solution: ..."

Use Newton's method to get an approximate value of x:

1. Change the equation to: $f(x)=3^x+5^x-12$ and solve the equation $f(x) = 0$ for x.

2. Newton's method:

$x_{n+1}=x_n - \frac{f(x_n)}{f'(x_n)}$

I assume that you know how to differentiate an exponential function(?).

3. Use $x_0 = 1$. After 4 steps of iteration you'll have the result given by Wolframalpha.

5. Sorry I am not as mathematically developed to appreciate a numerical solution as a hint.
Thank you though for your subsequent direction