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Mathematics 10 Online
OpenStudy (anonymous):

5^2x =3

OpenStudy (jdoe0001):

\(\bf 5^{2x} =3 \quad ?\)

OpenStudy (anonymous):

That's It

OpenStudy (jdoe0001):

have you covered logarithms yet?

OpenStudy (anonymous):

A little

OpenStudy (jdoe0001):

\(\bf 5^{2x} =3 \\ \quad \\ \textit{log cancellation rule of }log_{\color{red}{ a}}{\color{red}{ a}}^x=x\qquad thus \\ \quad \\ 5^{2x} =3\implies log_{{\color{brown}{ 5}}}({\color{brown}{ 5}}^{2x})=log_{{\color{brown}{ 5}}}3\implies 2x=log_{{\color{brown}{ 5}}}3\)

OpenStudy (jdoe0001):

then just solve for "x"

OpenStudy (anonymous):

2x3 + -9x2 + -5x = 0 Reorder the terms: -5x + -9x2 + 2x3 = 0 ... '(5 + -1x)' equal to zero and attempt to solve: Simplifying 5 + -1x = 0 Solving 5 + -1x = 0

OpenStudy (anonymous):

That's A Little Confusing Jdoe

OpenStudy (mathmale):

What is the objective of this problem? It's important that you include the instructions. Even more important is that you present the problem clearly enough so that potential helpers do not have to ask you for clarification. JDoe's method will certainly work. However, I think it easier to use either common logs (base 10) or natural logs (base e), unless the problem specifically asks for logs to the base 5.

OpenStudy (jdoe0001):

hmmm confusing?

OpenStudy (anonymous):

does my way help

OpenStudy (jdoe0001):

@LannaBoo24 it'd be if you haven't covered logarithms, yes

OpenStudy (mathmale):

@LannaBoo24 : Please indicate what you understand and what you do not understand here, so that either JDoe or I could step in with info to help you.

OpenStudy (anonymous):

how bout wat i said

OpenStudy (anonymous):

Well I Think The Base Of 10's Are Easier ThanThe Way Jdoe Said It. Heheheheh All Those Numbers Clumped Up Are Weird.

OpenStudy (anonymous):

The Objective Syas This Mathmale: Solve these equations. (Round decimal answers to the nearest hundredth.)

OpenStudy (anonymous):

Clarifies the difference between 'solving' and 'simplifying', and provides examples of ... 3 + 2x – 8 = –1 2x + 3 – 8 = –1 2x – 5 = –1 2x – 5 + 5 = –1 + 5 2x = 4 x = 2.

OpenStudy (mathmale):

@hehehehehehehe : Have you read this whole conversation all the way through? Lannaboo24 has verified that she meant \[\bf 5^{2x} =3\] so what is the relevance, if any, of your approach? @LannaBoo24 : Please take the common log of both sides of \[\bf 5^{2x} =3\]

OpenStudy (anonymous):

@mathmale calm down i just saw the question and helped i didn't see the comments

OpenStudy (anonymous):

Thank You Anyway Hehe.

OpenStudy (mathmale):

Perhaps you (hehe) should read through what others have typed before you comment.

OpenStudy (anonymous):

i don't have to

OpenStudy (anonymous):

Please Don't Start AN Argument On Here You Guys. And 5 &3 Have A Common log?

OpenStudy (mathmale):

@LannaBoo24 : Please take the common log of both sides of 5^(2x)=3 now.

OpenStudy (anonymous):

1.5?

OpenStudy (mathmale):

You could, if you wished, find the values of log 5 and log 3, but it's not strictly necessary. You should be applying rules of logarithms here.

OpenStudy (mathmale):

List the 3 rules of logs here, please. Hint: log ab = log a + log b

OpenStudy (anonymous):

Oh Okay Sorry I Was Thinking Common Factor

OpenStudy (mathmale):

Better look up and write down these rules of logs, as you'll need them later. You've got to have something to review if you want to learn these rules.

OpenStudy (mathmale):

@LannaBoo24 : Please take the common log of both sides of 5^(2x)=3 now.

OpenStudy (anonymous):

Log5= 0.69 and Log3=0.477

OpenStudy (mathmale):

True. But that's not what I asked you for, Lanna. Please take the common log of both sides of 5^(2x)=3 now. Note: You'll need to find, learn and apply the appropriate rule of logs to do this correctly. Again I ask you to list the three rules of logs used in problems such as this one. I gave you the first of these three rules, so that you'd know what I meant. What are the other two rules?

OpenStudy (mathmale):

If you have online notes, look up "rules of logs." If you don't, do an Internet search for "rules of logs."

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