please someone help me with logs. I am homeschooled an have no one else to talk to.?

Question

for example: log (base 4) 9x (squared) = log (base 2) (5x-8)

how would i solve this equation

Answer 1

This is how we write it on here... at least, to an extent. It helps to have a straightforward, easily read syntax.
log_{4} 9x² = log_{2} (5x-8)

Base conversion: log_b {a} = log_c {a} / log_c {b}
log_{2} (9x²) / log_{2} (4) = log_{2} (5x-8)

Simplify what you can, rearrange:
log_{2} (9x²) / 2 = log_{2} (5x-8)
log_{2} (9x²) = 2 ⋅ log_{2} (5x-8)

Exponent in a logarithm rule. log a^n = n log a
log_{2} (9x²) = log_{2} (5x-8)²

Remove logarithms
(9x²) = (5x-8)²
(3x)² = (5x-8)²
±(3x) = (5x-8)

3x = 5x - 8 ....... -3x = 5x - 8
2x = 8 .............. -8x = -8
x = 4 ................ x = 1

x = 1 is extraneous... only x = 4 works

Answer 2

I am not positive but I think this is how to do it.

rewrite 4 as 2 squared

(2^2)^9x^2 = 2^(5x-8)

2^ 18x^2 = 2^(5x-8)

18x^2 = 5x-8

solve
18x^2 - 5x + 8 =0
this cannot be factored so use the quadratic formula

Answer 3

since the bases are different you need to convert one log to the base of the other, using the change of base formula:
Log(baseb)M = log(new base)M/log(new base)b

Thus, log(base4) 9x^2 converts to
log(base2)9x^2 / log(base2)4 where log(base2)4 = 2

Now your equation is

[log(base2) 9x^2 ]/2 = log(base2) (5x - 8)

or
(1/2)[log(base2) 9x^2] = log base2) (5x - 8)
the 1/2 goes as a power on 9x^2

log(base2)(9x^2)^(1/2) = log(base2)(5x - 8)

the 1/2 pwer takes the square root
log(base2) 3x = log base2 (5x - 8)
since they are of same base, then the log numbers must equal

3x = 5x - 8
8 = 2x
4 = x

Answer 4

log b4 (9x²) = log b2 (5x - 8)
4 ^ (9x²) = 2 ^ (5x - 8)
(2²) ^ (9x²) = 2 ^ (5x - 8)
18x² = 5x - 8
18x² - 5x + 8 = 0
x = ( 5 ± 23.5 i ) / 36

Answer 5

What, your mom-teacher won't tell you? That doesn't mwake sense to me. You'd think you'd have an all-subject tutor if you were home schooled.