Log Equations / exponential equations

[kpx001]

I have three log equations about which I'm a bit confused. I need to solve for x.

1) 2 log₅(x) = 3 log₅(4)

My work: 2 log₅(x) - 3 log₅(4) = 0

The answer is "x = 8".

2) 3 log₂(x - 1) + log₂(4) = 5

My work: I tried doing (x-1)^3 and dividing the 3, but it doesn't work out.

The answer is "x = 3".

3) 3^1-2x = 4^x

I tried taking the natural log, but I can't get x by itself:

. . .(1 - 2x)ln(3) = xln(4)

Thank you!
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Edited by stapel -- Reason for edit: formatting

 

[galactus]

Try 'change of base'. One way.

\(\displaystyle \L\\3log_{2}(x-1)+log_{2}4=5\)

\(\displaystyle \L\\3\frac{log(x-1)}{log2}+\frac{log4}{log2}=5\)

\(\displaystyle \L\\3\frac{log(x-1)}{log2}+2=5\)

\(\displaystyle \L\\\frac{log(x-1)}{log2}=1\)

\(\displaystyle \L\\log(x-1)=log2\)

\(\displaystyle \L\\x-1=2\)

 

[stapel]

1) Use log rules to put the multipliers inside as powers:

. . . . .log₅(x²) = log₅(4³)

. . . . .log₅(x²) = log₅(64)

Once you have "log_b(something) = log_b(something else)", you can equate the arguments and solve.

2) Note that, since 32 = 2⁵, then 5 = log₂(32). Then the equation, after applying log rules, becomes:

. . . . .log₂((x - 1)³) = log₂(32) - log₂(4) = log₂(32/4) = log₂(8)

Equate the arguments and solve.

3) Taking the natural log should work. It's unfortunate that you didn't show your steps, so we can't help you correct your errors. You should have gotten:

. . . . .(1 - 2x)ln(3) = xln(4)

. . . . .ln(3) - x[2ln(3)] = x[ln(4)]

. . . . .ln(3) = x[ln(9) + ln(4)]

. . . . .ln(3) = x[ln(36)]

Divide through to isolate x.

Eliz.