## Alberta Free Tutoring And Homework Help For Math 30-1

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### If  $\log_8x=\frac{4}{3}y$log8x=43 y then the expression  $\log_2\left(64x^6\right)$log2(64x6) can be rewritten as:  $\log_2\left(64x^6\right)=a+by.$log2(64x6)=a+by. What are the values for a and b?

3 years ago log23 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

3 years ago 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 ( $\log_b\left(M^n\right)=n\log_bM$logb(Mn)=nlogbM ), 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$84y=x3  since -- if you recall from the beginning of the exponent and logs unit -- $\log_by=x$logby=x  is the inverse of  $x^b=y$xb=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$23·4y=x3

You can cancel out the powers of 3 by writing the cube roots of  $2^{3\cdot4y}$23·4y and  $x^3$x3  so that you get

$2^{4y}=x$24y=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  $\log_b\left(M\cdot N\right)=\log_bM+\log_bN$logb(M·N)=logbM+logbN .

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(24y)=(6·4y)log22 .

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

$a=6$a=6 and  $b=24$b=24 .

Hope this helps!

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

8 4/3 y=x

(23)4/3 y=x

24y=x

log2(64x6)

= log2(26x6)

=log2(2x)6

=6log2(2x)

=6(log22 + log2x)

=6(1+log2x) since log22=1

=6(1+log224y) substituting x=24y from above

=6(1+4ylog22)

=6+24y

so a=6 and b=24