Question

Solve the following equations for x

Answer 1

A. \[5=\log_{3}(x ^{2}+18)\]
B. \[-2=\log(3x+5) \]
C. \[e^{x}e^{(x+2)}=1\]

Answer 2

stupid logs I hate them

Answer 3

They are easy :) you just have to know the rules

Answer 4

yea me too!

Answer 5

Here's the rule that is the definition of log.

\( \log_b x = y \iff b^y = x\)

Apply this rule to your first equation. What do you get?

Answer 6

3^5=x^2

Answer 7

First off,

log(x) = y is the same as writing 10^y = x

Notation:

Remembering that log(x) is the lazy way of writing, \[\log_{10}(x)\], where 10 is just the base at which y is raised to.

Rules

Rules that apply to your problem:

If you put a log as the power to its base, it will cancel the logarithm out! So,

\[b^{\log_b(x)} = x\]
where b is just the base,

some examples of this rule in play

\[\log_{13}(x + 1) = y\]
\[13^{\log_{13}(x + 1)} = 13^{y}\]
\[(x + 1) = 13^{y}\]

Answer 8

Also I forgot to mention under Notation!

\[\ln(x) = \log_{e}(x)\]

Answer 9

The last question requires an exponent rule

Answer 10

let me show you give me a second to write it out, if you have any questions please post them

Answer 11

Exponent Rule

Two exponents can be added together if they have the same coefficient

\[13^{(x+1)}13^x = 13^{(x+1) + 13}\]

Answer 12

The rule is:
\(\log_b x = y \iff b^y = x\)

You have:

\(5 = \log_3 (x^2 + 18) \)

Change it into the form of the left side of the rule:

\(\log_3 (x^2 + 18) = 5\)

Now apply the rule:

\(\log_3 (x^2 + 18) = 5 \iff 3^5 = x^2 + 18\)

The right side is just a quadratic equation:

\( x^2 + 18 = 3^5\)

\(x ^2 + 18 = 343\)

\(x^2 = 225\)

\(x = \pm\sqrt{225}\)

\(x = \pm15\)

Answer 13

I guess I could write a general rule,

\[a^ba^c = a^{b+c}\]

where a, b and c are just numbers

Answer 14

@mathstudent55

I generally dont think you should give answers to these questions you should have given an unrelated example, as this will discourage cheating.

Also I just wanted to comment in your last part where you have,

\[x^2 = \pm \sqrt{255}\]

that is a correct solution but I dont think you properly explained the math behind it and why that is the answer, you should never make assumptions when teaching

Answer 15

how did you get 343 out of 3^5?

Answer 16

3 * 3 * 3 * 3 * 3 = 243.
The 343 is a typo.

Answer 17

Thanks for pointing it out.

Answer 18

Not to be a jerk or anything but it generally in the terms of service that we are not suppose to provide answers, not that I havent been guilty of it

Answer 19

x^2 = 255
\[x^{(2)\frac{1}{2}} = 255^\frac{1}{2}\]
\[x^{(\frac{2}{1})\frac{1}{2}} = 255^\frac{1}{2}\]
\[x^{\frac{(1)2}{2(1)}} = 255^\frac{1}{2}\]
\[x^{\frac{2}{2}} = 255^\frac{1}{2}\]
\[x^{1} = 255^\frac{1}{2}\]
\[x = \pm255^\frac{1}{2} = \pm \sqrt{255}\]

Answer 20

Note we use plus or minus because you always get positive numbers when any number is raised to the power of an even number.

for instance,

\[2^{2} = (2)2 = 2\]
\[(-2)^{2} = (-2)(-2) = 2\]
a negative multiplied by a negative gives you a positive

Answer 21

Just to note so you dont get confused you can divide any number by 1 and not change its value, I only did it to show the fraction multiplication.

\[x = \frac{x}{1}\]

Answer 22

If you get stuck or are having problems or are confused about what I'm talking about please tell me and I can help guide you to the answer

Answer 23

I think I got the answer to the #2 but could you check it for me?

Answer 24

I think you should write it like I have it written instead of using the arrow, that way you don't end up making a mistake. Although, the way you do it just seems awkward.

As for your question you should note this rule!

\[x^{-1} = \frac{1}{x}\]

Example,
\[5^{-3} = \frac{1}{5^3}\]
\[\frac{1}{4^{-3}} = 4^3\]


etc.

In algebra it is best to avoid decimals if you can help it, best to use your mind rather your calculator for algebra, you always want exact numbers so for 10^(-2), write it as

\[10^{-2} = \frac{1}{10^2}\]



so,

\[ 3x + 5 = \frac{1}{10^2}\]

At this point you should remember that what you can do to one side of the equaton you can do the other other.

\[4x + 3 = \frac{1}{20}\]
\[4x + 3 - 3 = \frac{1}{20} - 3\]
\[4x + 0 = \frac{1}{20} - 3\]
\[4 = \frac{1}{20} - 3\]
\[(4x)\frac{1}{4} = (\frac{1}{20} - 3)\frac{1}{4}\]
\[(\frac{4x}{1})\frac{1}{4} = (\frac{1}{20} - \frac{3}{1})\frac{1}{4}\]
\[\frac{4x}{4} = \frac{1}{20(4)} - \frac{3}{4}\]
\[\frac{4x}{4} = \frac{1}{20(4)} - \frac{3(20)}{4(20)}\]
\[x = \frac{1 - 3(20)}{4(20)} = \frac{1-60}{80} = \frac{-59}{80} = -\frac{59}{80}\]

The second last step I used the following rule,

\[\frac{a}{b} = \frac{a}{b}*\frac{n}{n} = \frac{an}{nb}, \frac{n}{n} = 1\]

for instance,

\[\frac{3}{2} = \frac{3}{2}*\frac{5}{5} = \frac{3*5}{2*5} = \frac{15}{10}\]

we can multiply it by 5/5 because 5/5 = 1, you can use division in this rule as well.

\[\frac{6}{4} = \frac{6}{4}*\frac{\frac{1}{2}}{\frac{1}{2}} =\frac{\frac{6}{2}}{\frac{4}{2}} = \frac{3}{2}\]

Answer 25

Does that help?

Answer 26

That helps! Thanks!