Question

Solve log11 (y + 8) + log11 4 = log11 60.

Answer 1

@phi

Answer 2

First, use this on the left side:
\(\log a + \log b = \log (ab)\)

Answer 3

Then use this:
If \(\log a = \log b\) then \(a = b\).

Answer 4

y=28 check right @mathstudent55

Answer 5

@mathstudent55 how do you solve this equation for answers between 0-2Pie \[3\tan ^{4}\alpha= 1+\sec ^{2}\alpha \]

Answer 6

y=28 check right

Answer 7

@Michele_Laino is it right check

Answer 8

for the trig equation i suggest you use the identity
sec^2 a = 1 + tan^2 a

Answer 9

y=28 am i right check

Answer 10

yes you are right with y=28

Answer 11

log 4(y + 8) = log 60
4y + 32 = 60

y = ?

Answer 12

not 28

Answer 13

4y+32=60
4y+32-32=60-32
4y=28
y=28 i'm right

Answer 14

@Michele_Laino i'm right see

Answer 15

4y = 28 not y

Answer 16

you forgot to divide by 4

Answer 17

y = 28/4

Answer 18

no fraction

Answer 19

please what is 28/4=...? @sallyfield888

Answer 20

\(\log_{11} (y + 8) + \log_{11} 4 = \log_{11} 60\)

\(\log_{11} (y + 8)4 = \log_{11} 60\)

\(\log_{11} (4y + 32) = \log_{11} 60\)

\(4y + 32 = 60\)

\(4y = 28\)

\(y = 7\)

Answer 21

7

Answer 22

@sallyfield888 Correct!

Answer 23

that's right! @sallyfield888

Answer 24

7 is right

Answer 25

3 tan^4 a = 1 + 1 + tan^2 a
3tan^4 a - tan^2 a - 2 = 0
you need to solve this equation

Answer 26

if you let t = tan^2 a
we get
3t^2 - t - 2 = 0

(3t + 2)(t - 1) = 0

so find t

t = tan^ a

Answer 27

from the values of tan ^ 2 a you can then find a

Answer 28

tan^2 a cannot be negative for real a so you cane ignore the negative root

Answer 29

im confused on how you went from the equatoin to the t =tan^2a

Answer 30

For this equation, let's look at a trig identity you need

\(3\tan ^{4}\alpha= 1+\sec ^{2}\alpha\)

Start here:
\(\sin^2 \alpha + \cos^2 \alpha = 1\)

\(\dfrac{\sin^2 \alpha}{\color{red}{\cos^2 \alpha}} + \dfrac{\cos^2 \alpha}{\color{red}{\cos^2 \alpha}} = \dfrac{1}{\color{red}{\cos^2 \alpha}}\)

\(\tan^2 \alpha + 1 = \sec^2 \alpha\)

Now use this substitution in your equation:

\(3\tan ^{4}\alpha= 1+\color{red}{\sec ^{2} \alpha}\)



\(3\tan ^{4}\alpha= 1+\color{red}{\tan^2 \alpha + 1}\)

\(3\tan ^{4}\alpha =\tan^2 \alpha + 2\)

\(3\tan ^{4}\alpha - \tan^2 \alpha - 2 = 0\)

Now you have a quadratic in \(tan^2 \alpha \)

You can solve it directly for \(\tan^2 \alpha\), or you can make a substitution.

Let's solve it for \(\tan^2 \alpha\) without making a substitution.

\((3\tan ^2\alpha + 2 )(\tan^2 \alpha - 1) = 0\)

\(3\tan ^2\alpha + 2 = 0\) or \(\tan^2 \alpha - 1 = 0\)

\(3\tan ^2\alpha = -2\) or \(\tan^2 \alpha = 1\)

Left equation gives no solution since a^2 = negative has no real solutions.

\(\tan^2 \alpha = 1\)

\(\tan \alpha = \pm \sqrt 1 \)

\(\tan \alpha = \pm 1\)

In degrees:

\(\alpha = 45^o \) or \(\alpha = 135^o\) or \(\alpha = 225^o \) or \(\alpha = 315^o\)

In radians:

\(\alpha = \dfrac{\pi}{4} \) or \(\alpha = \dfrac{3 \pi}{4}\) or \(\alpha = \dfrac{5 \pi}{4}\) or \(\alpha = \dfrac{7 \pi}{4}\)

Answer 31

thank you so much, that helped alot!!