Question

No idea how to solve?
Solve 4^x = 12

Answer 1

Hint:

\[\Large b^x = y\]

converts to

\[\Large \log_{b}(y) = x\]

Answer 2

okay...

Answer 3

does that make sense

Answer 4

or use \(b^x=A\iff x=\frac{\ln(A)}{\ln(b)}\) if you want a decimal approximation

Answer 5

Kinda, but could you maybe show me the steps you would take to get this answer. Sorry I grasp it better that way :/

Answer 6

Similar example:

Solve 5^x = 13 for x



\[\Large 5^x = 13\]


\[\Large \log_{5}(13) = x\]


\[\Large x = \log_{5}(13)\]


\[\Large x = \frac{\ln(13)}{\ln(5)}\]


\[\Large x \approx 1.59369264\]

Answer 7

That's not the answer to your problem, but it shows you how do these type of problems.

Answer 8

Got it thanks!

But what if its like: 3^x – 4 = 7^x + 9

Answer 9

are x-4 and x+9 exponents?

Answer 10

Yes!

Answer 11

is the problem

\[\Large 3^{x-4} = 7^{x+9}\]

Answer 12

Yes :)

Answer 13

alright, one moment

Answer 14

Similar Example:


Solve \(\Large 7^{x-2} = 5^{x+6}\) for x


\[\Large 7^{x-2} = 5^{x+6}\]


\[\Large \ln\left(7^{x-2}\right) = \ln\left(5^{x+6}\right)\]


\[\Large (x-2)\ln(7) = (x+6)\ln(5)\]


\[\Large x*\ln(7)-2\ln(7) = x*\ln(5)+6\ln(5)\]


\[\Large x*\ln(7) = x*\ln(5)+6\ln(5)+2\ln(7)\]


\[\Large x*\ln(7)-x*\ln(5) = 6\ln(5)+2\ln(7)\]


\[\Large x\left(\ln(7)-\ln(5)\right) = 6\ln(5)+2\ln(7)\]


\[\Large x = \frac{6\ln(5)+2\ln(7)}{\ln(7)-\ln(5)}\]


\[\Large x \approx 40.2661684921349\]


Note: the last line is the approximate solution and the second to last line is the exact solution


This isn't the answer to your problem, but it shows you how to solve problems like this.

Answer 15

Thanks!

Answer 16

you're welcome