Question

log base 4 (x^2-3) + log base 4 10= 1

Answer 1

log(base4) (x^2 - 3) + log(base 4) 10 = 1
addition of logs is the same as multiply so we can write it:
log4(10*(x^2-3)) = 1
log4(10x^2 - 30) = 1
Write the exponent equiv of logs
4^1 = 10x^2 - 30
4 = 10x^2 - 30
4 + 30 = 10x^2
34 = 10x^2
x^2 = 34/10
x^2 = 3.4
x = underroot 3.4
x = 1.844
:
2) log(base 5)3 - log(base 5) 5x = 2
subtracting logs is the same as divide so we can write it
log5(3/5x)2
exponent equiv
5^2 = 3/5x
25 = 3/5x
Multiply by 5x
25(5x) = 3
125x = 3
x = 3/135

Answer 2

the answer is +/- square root 85/5 how do I get there?

Answer 3

product rule for logarithms
\[\log_{a}x+\log_{a}y=\log_{a}xy\]
\[\log_{4}[10(x^2-3)]=1\]
convert to exponential notation
\[4^1=10(x^2-3)\]
\[10x^2-30=4\]
\[10x^2-34=0\]
\[5x^2-17=0\]
Now you can apply the quadratic formula or complete the square to solve for x

Answer 4

start with the left hand side

\[\log _{4} (x^2 - 3) + \log _{4} 10 = \log _{4} (10(x^2 - 3))\]

raising each term to be a power of 4

10x^2 - 30 = 4

10x^2 = 34

x^2 = 34/10

\[x = \pm \sqrt{34/10}\]

Answer 5

@ all my ans.. is correct

Answer 6

x = sqrt (3.4) = 1.84

Answer 7

the answer is +/- square root 85/5 how do I get there?

Answer 8

\[\log_4 (x^2-3)+ \log_4 (10)=1\]
We know that
\[\log_a b+ \log_a c=\log_a bc\]
so using this here
\[\log_4 (x^2-3) 10=1\]
The most fundamental property of logarithms
\[\log_a b=c=>b=a^c\]
Using this we get
\[(x^2-3) \times 10=4^1\]
We get
\[10x^2-30=4\]
or
\[10x^2=34\]
\[x^2=\frac{34}{10}\]
\[x^2=\frac{17}{5}\]
Getting to your answer is quite easy
Multiply numerator and denominator by 5
\[x^2=\frac{17 \times 5}{5 \times 5}\]
We get
\[x^2=\frac{85}{25}\]
Take root both sides we get
\[x=\pm \sqrt{\frac{85}{25}}\]
we get finally
\[x=\pm \frac{\sqrt{85}}{5}\]