logarithms

[Guest]

1) 3log(5x)+4log(2x)=3

These are the steps I did:
log10[(5x^3)(2x^4)]=3
log10[(125x^3)(16x^4)]=3
log10=2000x^7=3
10^3=2000x^7

--I tried diving each side by 1000, since 10^3=1000, and then multiplying 1/7 because of the x^7, but the answer was wrong. My book says the correct answer is .9057. I don't understand how they found that.


2) log2(x^8)=7

I found 2^7=x^8, and 128=x^8

--My book said the correct answer is 1.8340, and I don't understand how they got that.

 

[stapel]

Your formatting is ambiguous, but I'm pretty sure your first and second lines (in your solution) are okay. But what do you mean by "log₁₀="?

If that's a typo, then I think the next line is okay. But, given:

. . . . .1000 = 2000x⁷

...why would you divide by 1000 instead of by 2000 (to isolate the variable)? And why would you multiply x⁷ by 1/7?

I'm sorry, but I'm not following your reasoning, and/or I'm not understanding your formatting. Shouldn't you divide through by 2000 and then take seventh roots of either side?

Eliz.

 

[Guest]

Eliz, My books says to multiply 1/7 on each side because of the x^7.

 

[stapel]

But the steps you've listed would give you "1/7 = (1000/7)x⁷", not a solution for x.

Sorry, but I don't know what your book is getting at. I can only tell you what would actually solve the equation.

Eliz.

 

[soroban]

Hello, Angela!

2) log₂(x⁸) = 7

I found 2⁷ = x⁸, and 128 = x⁸ . . . . right!

My book said the correct answer is 1.8340
and I don't understand how they got that.

You're one step from the answer . . .

We have: . x⁸ .= .128

How do we solve for x? . We take the 8^th root of both sides.

. . and we have: . x .= .128^1/8 .= .1.834008086...

 

[soroban]

Hello, Angela!

1) 3 log(5x) + 4 log(2x) = 3

I wish I could see what you book did . . . that 1/7 is weird!

We have: . log(5x)³ + log(2x)⁴ .= .3

. . . . . . log(125x³) + log(16x⁴) .= .3

. . . . . . . . . . . log(125x³·16x⁴) .= .3

. . . . . . . . . . . . . . log(2000x⁷) .= .3

Then: . . . . . . . . . . . . .2000x⁷ .= .10³ .= .1000

. . . . . . . . . . . . . . . . . . . . . .x⁷ .= .0.5

. . . . . . . . . . . . . . . . . . . . . . x .= .(0.5)^1/7 .= .0.905723664

 

[soroban]

Hello, all!

I think I know what the book did . . . a really stupid move . . .

It got to: .. . . log(2000x⁷) .= .3

Then: .log(2000) + log(x⁷) .= .3

. . . . . . . . . . . . . . . .log(x⁷) .= .3 - log(2000)

. . . . . . . . . . . . . . . 7·log(x) .= .3 - log(2000)

And now "multiply by 1/7" makes sense
. . but it's still a very stupid method.

Solve for x and we get this grotesque answer: .x .= .10^{[3 - log(2000)]/7}

. . See what I mean? . . . really truly STUPID!
. . (Authors like that will bring back public flogging.)