Answered By Adham R

log_2³ X= 4/3y

log₂ X= 4/3y*3=4y

log₂(64x⁶)=log₂(64)+log₂(x⁶)=log₂(64)+6*log₂(x)=log₂(64)+6*4y=6+24y

a=6

b=24

Answered By Sophia E

You have two equations to deal with:

1)  $\log_8x=\frac{4}{3}y$log₈x=43 y 

2)  $\log_2\left(64x^6\right)=a+by$log₂(64x⁶)=a+by 

First, you want to look at Equation #1. Multiply both sides by 3 so that you can get rid of the denominator on the right-hand side. Using the log laws by multiplying the exponent ( $\log_b\left(M^n\right)=n\log_bM$log_b(Mⁿ)=nlog_bM ), you can change the left hand side of the equation to

 $3\log_8x=4y$3log₈x=4y  

  $\log_8x^3=4y$log₈x³=4y  .

To find x from the first equation, you can rearrange the equation to

 $8^{4y}=x^3$8^4y=x³  since -- if you recall from the beginning of the exponent and logs unit -- $\log_by=x$log_by=x  is the inverse of  $x^b=y$x^b=y .

8 is the same as  $2^3$2³ , so you can rewrite the left-hand side of the new equation again to raise both sides to the third power.

 $2^{3\cdot4y}=x^3$2^3·4y=x³ 

You can cancel out the powers of 3 by writing the cube roots of  $2^{3\cdot4y}$2^3·4y and  $x^3$x³  so that you get

 $2^{4y}=x$2^4y=‍x . Now that you found x, you can plug this into the second equation.

Now, look at the second equation

  $\log_2\left(64x^6\right)=a+by$log₂(64x⁶)=a+by .

You can rewrite the left-hand side as

 $\log_264+\log_2x^6=a+by$log₂64+log₂x⁶=a+by  using the log law for which  $\log_b\left(M\cdot N\right)=\log_bM+\log_bN$log_b(M·N)=log_bM+log_bN .

Furthermore, you can rewrite  $\log_2x^6$log₂x⁶ as  $6\log_2x$6log₂x .

 $\log_264=6$log₂64=6 and

 $6\log_2x=6\log_2\left(2^{4y}\right)=\left(6\cdot4y\right)\log_22$6log₂x=6log₂(2^4y)=(6·4y)log₂2 .

 $\log_22=1$log₂2=1 , so you're left with  $6\cdot4y=24y$6·4y=24y .

Your final answer for the values of a and b are:

 $a=6$a=6 and  $b=24$b=24 .

Hope this helps!

Answered By Becky L

Change log₈x=$\frac{4}{3}y$43 y to exponential function, we have

8 ^{4/3 y}=x

(2³)^{4/3 y}=x

2^4y=x

log₂(64x⁶)

= log₂(2⁶x⁶)

=log₂(2x)⁶

=6log₂(2x)

=6(log₂2 + log₂x)

=6(1+log₂x) since log₂2=1

=6(1+log₂2^4y) substituting x=2^4y from above

=6(1+4ylog₂2)

=6+24y

so a=6 and b=24