/sci/ - Science & Math

[Anonymous No.2072246]

I need some help with a type of problem in which no explanation was given in math class

please help me guys
we're doing a chapter about logarithms
6^(9x+2=30) solve for x (please show me work, I just want to know how this works so I can apply it to the rest of my work)

[Anonymous No.2072263]

BUMP

[Anonymous No.2072273]

please, i just need help on one problem and my teacher left it completely out of the lesson plan. Without knowing how to answer at least 1 of these online, i will be unable to take a quiz on a unit i completely understand otherwise

[Anonymous No.2072281]

Easy:
<span class="math">
\begin{align}
6^{9x+2}&=30 \\
\ln 6^{9x+2}&=\ln 30 \\
(9x+2)\ln 6&= \ln 6 + \ln 5 \\
9x + 2 &= \frac{\ln 6 + \ln 5}{\ln 6} \\
x &= \frac{\frac{\ln 5}{\ln 6} - 1 }{3}
\end{align}
[/spoiler]

[Anonymous No.2072291]

Repost due to poor latex formatting:
<span class="math">
\begin{math}
6^{9x+2}&=30
\ln 6^{9x+2}&=\ln 30
(9x+2)\ln 6&= \ln 6 + \ln 5
9x + 2 &= \frac{\ln 6 + \ln 5}{\ln 6}
x &= \frac{\frac{\ln 5}{\ln 6} - 1 }{3}
\end{math}
[/spoiler]

[Anonymous No.2072292]

MFW>>2072281
I just realized i fucked up the problem
6^(9x+2)=30 solve for x
this is how it should have been

[Anonymous No.2072296]

Oh for fuck sake...!
<span class="math">
6^{9x+2}&=30 \ln 6^{9x+2}&=\ln 30 (9x+2)\ln 6&= \ln 6 + \ln 5 9x + 2 &= \frac{\ln 6 + \ln 5}{\ln 6} x &= \frac{\frac{\ln 5}{\ln 6} - 1 }{3} [/spoiler]

[Anonymous No.2072303]

<span class="math">
6^{9x+2}=30
\ln 6^{9x+2}=\ln 30
(9x+2)\ln 6= \ln 6 + \ln 5
9x + 2 = \frac{\ln 6 + \ln 5}{\ln 6}
x = \frac{\frac{\ln 5}{\ln 6} - 1 }{3}
[/spoiler]

[Anonymous No.2072308]

<span class="math">
6^{9x+2}=30 \\
\ln 6^{9x+2}=\ln 30
(9x+2)\ln 6= \ln 6 + \ln 5 \\
9x + 2 = \frac{\ln 6 + \ln 5}{\ln 6} \\
x = \frac{\frac{\ln 5}{\ln 6} - 1 }{3}
[/spoiler]

[Anonymous No.2072310]

spacings bruh, also, what do you write latex with?

[Anonymous No.2072316]

<span class="math">
6^{9x+2}=30; \ln 6^{9x+2}=\ln 30 ; (9x+2)\ln 6= \ln 6 + \ln 5; 9x + 2 = \frac{\ln 6 + \ln 5}{\ln 6}; x = \frac{\frac{\ln 5}{\ln 6} - 1 }{3} \qedhere
[/spoiler]

[Anonymous No.2072318]

ok
so
30 = 6^(9x + 2)
divide both sides by 6
5 = 1^(9x + 2)
but
5/9x+2 = 1
9x + 2 = 5
subtract 2
7x = 3
7/3x = 1
73x/ = 7
x/ = 7/73
x = 773

[Anonymous No.2072327]

Use the [ math ] tags. I'm used to a more tolerant flavor of LaTeX, doesn't seem like e.g. align-environments work here.
<span class="math">
6^{9x+2}=30; \ln 6^{9x+2}=\ln 30 ; (9x+2)\ln 6= \ln 6 + \ln 5; 9x + 2 = \frac{\ln 6 + \ln 5}{\ln 6}; x = \frac{\frac{\ln 5}{\ln 6} - 1 }{3}
[/spoiler]

[Anonymous No.2072329]

>>2072318
You win at math. If only I could have figured out how to preform algebraic operations on transcendental functions before you I might have had a shot at the Nobel prize.

Oh well.

[Anonymous No.2072347]

Protip: WolframAlpha knows all.
http://www.wolframalpha.com/input/?i=6^(9x%2B2)%3D30

[Anonymous No.2072459]

Assume you can use calculator:

6^(9x+2) = 30
(9x + 2)log(6) = log(30)
9x + 2 = log(30)/log(6)
9x = log(30)/log(6) - 2
x = [log(30)/log(6) - 2] / 9

x = -.0113