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Solve 3(7^x+1) = 10 -2(7^x+1) I cant figure out what to do after the first step
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Is it 3(7^(x+1)) or 3(7^(x)+1?
In other words, is the +1 in the exponent?
x+1 is the exponent, 7 is not
\[3(7^x+1) = 10 -2(7^x+1) \] add 2(7^x+1) to both sides \[5(7^x+1)=10\] divide both sides by 5 \[7^x+1=2\] -1 to both sides \[7^x=1\] \[\large \log_7(7^x)=\log_7 (1)\] \[\large \cancel{\log_7}(\cancel{7}^x)=\log_7 (1)\] \[x=\log_7(1)\] \[x=0\]
oh blah didnt read the bit about the exponent until now
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:( sorry sometimes it is difficult to write the problem as is
\[3(7^{x+1})=10-2(7^{x+1})\] \[5(7^{x+1})=10\] \[7^{x+1}=2\] \[\log_7 7^{x+1}=\log_7 2\] \[x+1=\log_7 2\] \[x=\log_7 (2)-1\]
Thank you!! this is very helpful
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