2^x+3=5^x If someone could take me step by step in this problem????
okay, so in this problem, you have to be familiar with the log function (which i assume you are)
A little, thats what we are learning now
I know that I have to natural (ln) each side..
I came up with an answer of 2.269 I just know its wrong...
if someone is good with vectors here, so plz give some thought to my question as well.
sorry, logs are so far back in my math knowledge. im trying to remember the rules
\[2^x+3=5^x\] or \[2^{x+3}=5^x\]
Here is my work... maybe someone can check it to see where I went wrong. 2^(x+3)=5^x Then I divided each side by x ln 2 and it left me with: x+3/x=ln5/ln2
The second one...
ohhh i remember how to do it. i thought it was 2^x + 3 instead of 2^(x+3)
I dont know how to write equations on here.... so sorry
its cool. alright, so log both sides to start. that will get you ln(2^(x+3)) = ln(5^x)
then with the logarithm rules, you can bring the exponent of both sides in front. this gives you: (x+3)ln(2) = xln(5)
then you switch things around to get (x+3)/x = ln(5)/ln(2)
Ok, I got that. So far, so good
then you can split the fraction on the left up to make two different ones. It becomes (x/x) + (3/x)
or, in other words, (3/x) + 1
OK... right here is where my teacher told us to croos multiply?
cross sorry. So it gave me x+3/x=2.322/1
does your teacher want you to get an exact answer? or are decimals okay
Decimals to the thousandth
okay, well i would highly recommend solving for x in exact form first, and then getting the decimal. it's much more pure math. if you have (x/3) + 1 = ln(5)/ln(2) it's very easy to get it to exact form
sorry, (3/x)
just move the 1 over to the right side
What one? where are you getting the one from?
then it will be 3/((ln(5)/ln(2))+1)
from the left side. where it says (3/x) + 1 = ln(5)/ln(2)
it becomes 3/x = ln(5)/ln(2) -1
which becomes 3/((ln(5)/ln(2))-1)
^^ that is equal to x
is the answer 1.689?
i believe its 2.269. make sure you plug it in correctly on your calculator.
Thats what I had for my answer to begin with?
So I am right?
haha, yes
Thanks for your help.
no problem
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