Exponential and logarithmic Equation Problems

[onemic]

Homework Statement

Evaluate each of the following expressions without using a calculator.

1) log₂16√8Solve for the unknown value in each of the following equations without using a calculator.

2) 3^(x+4)−5(3^x)=684

3) 7(4^2x)=28(4^x)

Homework Equations

Exponent law for multiplication

The Attempt at a Solution

1)

log₂(16)√8
=log₂32√2
=log₂32^1/2

Im not really sure what to do at this point or if my approach is just completely wrong

2)

3^(x+4)−5(3^x)=684
=(3^x)(3⁴)-5(3^x)=684
Let a = 3^x
(a)(3⁴)-5a=684

Once again I am not sure if my approach is completely wrong or I am just at a loss on what to do afterward.

3)

7(4^2x)=28(4^x)
7(4^x)²=28(4^x)
Let a=4^x
7a²=28a
a²=28a/7
a²=4a

Im not sure what to do after this

 

[Buffu]

$\log_2 ((16)\sqrt{8})$ or $\sqrt{8}\times \log_2 ((16))$ ?

 

[onemic]

$\log_2 ((16)\sqrt{8})$ or $\sqrt{8}\times \log_2 ((16))$ ?

The former

I edited the OP to make it a little clearer

 

[Buffu]

The former

I edited the OP to make it a little clearer

Use $\log_a(bc) = \log_a(b) + \log_a(c)$, $16 = 2^4$ and same for $\sqrt{8}$.

For the other two you need to take log on both sides.

 

[onemic]

so the first one would be:

$\log_2(16) + \log_2(\sqrt{8})$
$\log_2(2^4) + \log_2(2\sqrt{2})$

and I am not really sure what to do with the square root

For the other two questions, is what I have so far correct and I need to take the log on both sides from the point where I am at or is the whole thing wrong?

 

[Buffu]

so the first one would be:

$\log_2(16) + \log_2(\sqrt{8})$
$\log_2(2^4) + \log_2(2\sqrt{2})$

and I am not really sure what to do with the square root

For the other two questions, is what I have so far correct and I need to take the log on both sides from the point where I am at or is the whole thing wrong?

No you are not getting it. You should try to use $\log_a(b^c) =c\log_ab$. Now if $a =b$ then $c\log_a(a) = ?$

For the other two questions, is what I have so far correct and I need to take the log on both sides from the point where I am at or is the whole thing wrong?

What you did is correct, now you need to solve for 'a' then take log to 'x'.

 

[ehild]

so the first one would be:

$\log_2(16) + \log_2(\sqrt{8})$
##\log_2(2^4) + \log_2(2\sqrt{2})##

Write $\sqrt{8}$ in form of 2^a, power of 2 .

 

[ehild]

2)

3^(x+4)−5(3^x)=684
=(3^x)(3⁴)-5(3^x)=684
Let a = 3^x
(a)(3⁴)-5a=684

Once again I am not sure if my approach is completely wrong or I am just at a loss on what to do afterward.

How much is 3⁴?

3)

7(4^2x)=28(4^x)
7(4^x)²=28(4^x)
Let a=4^x
7a²=28a
a²=28a/7
a²=4a

Im not sure what to do after this

Rewrite the equation in form a²-4a =0, and factor out a. Remember, a product is zero if any of the factors is zero. What solutions do you get for a? Are both valid?

 

[onemic]

Sorry for the super late reply. I decided to continue on with other questions and the next unit before coming back to this.
1)

$\log_2(16) + \log_2(\sqrt{8})$
$\log_2(2^4)(2^3)^{0.5}$
$\log_2(2^4)(2^{1.5})$
$\log_2(2^{5.5})$
$5.5\log_2(2)$
$=5.5$

2)

wow I don't know why this question shook me. The answer was in my face the entire time haha.
(starting from where I left off in the OP)
$81a-5a=684$
$76a=684$
$a=684/76$
$a=9$
$3^x=9$
$x=2$

3)
(starting from where I left off in the OP)
$a^2=4a$
$4^{2x}=(4)4^x$
$4^{2x}=4^{x+1}$
$2x=x+1$
$2x-x=1$
$x=1$

EDIT: thanks for the heads up on the typo

 

[Buffu]

Sorry for the super late reply. I decided to continue on with other questions and the next unit before coming back to this.
1)

$\log_2(16) + \log_2(\sqrt{8})$
$\log_2(2^4)(2^3)^\color{red}{0.5}$
$\log_2(2^4)(2^{1.5})$
$\log_2(2^{5.5})$
$5.5\log_2(2)$
$=5.5$

2)

wow I don't know why this question shook me. The answer was in my face the entire time haha.
(starting from where I left off in the OP)
$81a-5a=684$
$76a=684$
$a=684/76$
$a=9$
$3^x=9$
$x=2$

3)
(starting from where I left off in the OP)
$a^2=4a$
$4^{2x}=(4)4^x$
$4^{2x}=4^{x+1}$
$2x=x+1$
$2x-x=1$
$x=1$

I guess it is correct except one typo.

 

[Mark44]

(starting from where I left off in the OP)
$81a-5a=684$
$76a=684$
$a=684/76$
$a=9$
$3^x=9$
$x=2$

It's a good habit to get into to check your answer by substituting your value of x in the original equation.

3)
(starting from where I left off in the OP)
$a^2=4a$
$4^{2x}=(4)4^x$
$4^{2x}=4^{x+1}$
$2x=x+1$
$2x-x=1$
$x=1$

You started off in the direction that @ehild suggested, but then went off in a different direction. Following the direction she suggested you would have this:
$4^{2x} - 4\cdot4^x = 0$
$4^x(4^x - 4) = 0$
Are you sure there aren't solutions other than x = 1, which is the one you showed?

 

[onemic]

It's a good habit to get into to check your answer by substituting your value of x in the original equation.

You started off in the direction that @ehild suggested, but then went off in a different direction. Following the direction she suggested you would have this:
$4^{2x} - 4\cdot4^x = 0$
$4^x(4^x - 4) = 0$
Are you sure there aren't solutions other than x = 1, which is the one you showed?

doing it that way I get:

$a^2-4a=0$
$a(a-4)=0$
$a=4$ or $a=0$

$a=4$
$4^x=4$
$x=1$

$a=0$
$4^x=0$
*no solutions*

therefore the correct answer is $x=1$

 

[Mark44]

doing it that way I get:

$a^2-4a=0$
$a(a-4)=0$
$a=4$ or $a=0$

$a=4$
$4^x=4$
$x=1$

$a=0$
$4^x=0$
*no solutions*

therefore the correct answer is $x=1$

Looks good!