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need some algebra help
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x = ln(5/8) x = ln(5) - ln(8)
do you know how to put it into log form? i would first put this in exponent form and work through trying to get x alone.
\[\log_{\frac{ 1 }{ 4}}32 = x+1\]
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I will continue this TOMMOROW do not close this mods
@saltysuga wrote:
x = ln(5/8)
x = ln(5) - ln(8)
@ykbatman wrote:
@saltysuga wrote:
x = ln(5/8)
x = ln(5) - ln(8)
\[(\frac{ 1 }{ 4 })^{x+1}=32\] \[\left( \frac{ 1 }{ 2^2 } \right)^{x+1}=2^{5}\] \[\left( 1\times2^{-2} \right)^{x+1}=2^{5}\] \[\left( 2 \right)^{-2(x+1)}=2^{5}\] -2(x+1)=5 -2x-2=5 -2x=5+2 -2x=7 \[x=-\frac{ 7 }{ 2 }\]
crumb crumingtion never did his so called "continuing" on this question ;-;
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@ykbatman wrote:
@saltysuga wrote:
x = ln(5/8)
x = ln(5) - ln(8)
@crumbcrumbington wrote:
@ykbatman wrote:
@saltysuga wrote:
x = ln(5/8)
x = ln(5) - ln(8)
@adrianluvvsyouu2 wrote:
@crumbcrumbington wrote:
@ykbatman wrote:
@saltysuga wrote:
x = ln(5/8)
x = ln(5) - ln(8)
@adrianluvvsyouu2 wrote:
@crumbcrumbington wrote:
@ykbatman wrote:
@saltysuga wrote:
x = ln(5/8)
x = ln(5) - ln(8)
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